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Full text of "Engineering thermodynamics"

.ENGINEERING 
THERMODYNAMICS 



BY 

CHARLES EDWARD LUCKE, PH.D., - 

f> $ 

Professor of Mechanical Engineering in Columbia University, New York City 



McGEAW-HILL BOOK COMPANY 
239 WEST 39TH STREET, NEW YORK 

6 BOUVERIE STREET, LONDON, E. C. 
1912 




PREFACE 



CALCULATIONS about heat as a form of energy, and about work, another 
related form, both of them in connection with changes in the condition of all 
sorts of substances that may give or take heat, and perform or receive work 
while changing condition, constitute the subject matter of this book. The 
treatment of the subject matter of this text is the result of personal experience 
in professional engineering practice and teaching students of engineering at 
Columbia University. 

Even a brief examination of the conditions surrounding changes in sub- 
stances as they gain or lose heat, do work or have work done on them, and of 
the corresponding relations between heat and work as forms of energy independ- 
ent of substances, will convince any one that the subject is one of great com- 
plexity. Accordingly the simplicity needed for practical use in the industries 
can be reached only by a consideration of a great mass of sub-topics and data. 
That the doing of work, and the changes in heat content of substances were 
related phenomena, and that these relations when formulated, would con- 
stitute a branch of science, was conceived about a half century ago, and the 
science was named Thermodynamics. The Engineer Rankine, who helped 
to create it, defined thermodynamics as " the reduction of the laws according 
to which such phenomena took place to a physical theory or corrected system 
of principles." Since Rankine's time thermodynamics has become a very 
highly developed science and has proved of great assistance in the formu- 

| lation of modern physical chemistry, and to those branches of engineering 
that are concerned with heat. Unfortunately, as thermodynamics developed 

[ as a separate subject it did not render proportionate service to engineering, 

i which itself developed even more rapidly in the same period under the guidance 
of men whose duty it was to create industrial apparatus and make it work 
properly, and who had little or no time to keep in touch with purely scientific 

: advances or to interpret such advances for utilitarian ends. Thermodynamics 
proper is concerned with no numerical quantities nor with any particular 
substance nor for that matter with any actual substances whatever, but it is a 
physical theory of energy in relation to matter as a branch of natural philosophy. 
Engineering, however, is concerned with real substances, such as coal, steam, 
arid gases and with numerical quantities, horse-powers of engines, 'temper- 
atures of steam, the heats of combustion of oils, so that alone, the principles 
of thermodynamic philosophy will not yield a solution of a practical problem, 

v 

271310 



vi PREFACE 

be it one of design or one of analysis of test performance of actual heat machine 
or thermal apparatus. It is the province of engineering thermodynamics to 
guide numerical computation on thermal problems for real substances being 
treated in real apparatus. Its field, while including some of that of pure 
thermodynamics, extends far beyond the established provinces of that subject 
and extends to the interpretation of all pertinent principles and facts for purely 
useful purposes. Engineering thermodynamics, while using whatever prin- 
ciples of pure thermodynamics may help to solve its problems, must rely on 
a great mass of facts or relations that may not yet have risen to the dignity 
of thermodynamic laws. The workers in shops, factories, power plants or 
laboratories engaged in designing or operating to the best advantage machines 
and apparatus using heat with all sorts of substances, have developed great 
quantities of rules, methods and data that directly contribute to the ends sought. 
While for each class or type of apparatus there has grown up a separate set 
of data and methods in which much is common to several or all groups, not 
nearly so much assistance is rendered by one to another as should be by a proper 
use of engineering thermodynamics, which applies methods, principles and 
conclusion to all related problems. Classes of apparatus about which such 
groups of methods of analysis or synthesis, or collections of special data 
have developed and which it is province of engineering thermodynamics to 
unify so far as may be, include air compressors, and compressed air engines, 
reciprocating steam engines, steam turbines, steam boilers, coal-, oil- and gas- 
fired furnaces, gasifiers of coal and oil, gas producers, gas engines, complete 
steam or gas power plants, mechanical refrigeration and ice-making plants and 
chemical factory equipment, or more generally, machinery and apparatus for 
heating and cooling, evaporating and condensing, melting and freezing, moisten- 
ing and drying, gasification and combustiom. 

The nature of the subject and its division are better indicated by the 
classes of problems to be solved by its aid or the contributions expected of it 
than by the kinds of apparatus to which they apply. Probably its broadest 
contribution is the establishment of limits of possible performance of heat 
apparatus and machines. These limits will show what might be expected of 
a steam engine, gas engine or refrigerating machine when its mechanism is quite 
perfect and thus they become standards of reference with which actual per- 
formance can be compared, and a measure of the improvements yet possible. 
These same methods and practices are applicable to the analysis of the operat- 
ing performance of separate units and complete plants to discover the amount 
of energy being lost, how the total amount is divided between the different 
elements of the apparatus, which of the losses can be prevented and how, and 
finally which are absolutely unavoidable. This sort of analysis of the per- 
formance of thermal apparatus is the first step to be taken by the designer 
or manufacturer to improve the machine that he is creating for sale, and is 
essential to the purchaser and user of the machine, who cannot possibly keep 
it in the best operating condition without continually analyzing its performance 
and comparing results with thermodynamic possibilities. 



PREFACE vii 

The subject naturally divides into three general parts, the first dealing 
with the conditions surrounding the doing of work without any consideration 
of heat changes, the second heat gains and losses by substances without reference 
to work involved and the third, transformation of heat into work or work into 
heat in conjunction with changes in the condition of substances. The first 
part applies to the behavior of fluids in the cylinders of compressors and engines. 
The second part is concerned with the development of heat by combustion, 
its transmission from place to place, and the effect on the physical condition of 
solids, liquids, gases with their mixtures, solutions and reactions. The third 
part is fundamental to the efficient production of power by gases in internal 
combustion gas engines or compressed-air engines and by steam or other vapors 
in steam engines and turbines, and likewise as well to the production of 
mechanical refrigeration by ammonia, carbon dioxide and other vapors. 

Accordingly, the six chapters of the book treat these three parts in order. 
The first three chapters deal with work without any particular reference to 
heat, the second two with heat, without any particular reference to work, 
while the last is concerned with the relation between heat and work. After 
establishing in the first chapter the necessary units and basic principles for 
fixing quantities of work, the second chapter proceeds at once to the determina- 
tion of the work done in compressor cylinders and the third, the available work 
in engine cylinders, in terms of all the different variables that may determine 
the work for given dimensions of cylinder or for given quantities of fluid. There 
is established in these first three chapters a series of formulas directly applicable 
to a great variety of circumstances met with in ordinary practice. All are 
derived from a few simple principles and left in such form as to be readily 
available for numerical substitution. This permits of the solution of numerical 
problems on engine and compressor horse-power, fluid consumption or capacity 
with very little labor or time, although it has required the expansion of the 
subject over a considerable number of pages of book matter. A similar pro- 
cedure is followed in the succeeding chapters, formulas and data are developed 
and placed always with a view to the maximum clearness and utility. The 
essential unity of the entire subject has been preserved in that all the important 
related subjects are treated in the same consistent manner and at sufficient 
length to make them clear. When no general principles were available for 
a particular solution there has been no hesitation in reverting to specific data. 
The subject could have been treated in a very much smaller space with less 
labor in book writing but necessitating far greater labor in numerical work 
on the part of the user. This same aim, that is, the saving of the user's 
time and facilitating the arrival at numerical answers, is responsible for the 
insertion of a very considerable number of large tables, numerous original 
diagrams and charts, all calculated for . the purpose and drawn to scale. 
These, however, take a great deal of room but are so extremely useful in 
everyday work as to justify any amount of space thus taken up. For 
the sake of clearness all the steps in the derivation of any formula used 
arc given, and numerical examples are added to illustrate their meaning 



viii PREFACE 

and application, This also requires a considerable amount of space but 
out it the limitations of the formulas would never be clear nor could a student 
learn the subject without material assistance. Similarly, space has been used 
in many parts of the book by writing formulas out in words instead of express- 
ing them in symbols. This saves a great deal of time and labor in hunting up 
the meaning of symbols by one who desires to use an unfamiliar formula involv- 
ing complex quantities, the meaning of which is often not clear when it is entirely 
symbolic. Thus, in the discussion of steam boiler capacity and efficiency, a 
dozen or more pages are taken up with formulas that could have been con- 
centrated in a single page were symbols used entirely, but only at the sacrifice 
of clearness and utility. Where in the derivation of a new formula or in the 
treatment of a new subject, reference to an old formula or statement is needed 
and important, repetition is resorted to, rather than mere reference, so that 
the new topic may be clear where presented, without constantly turning the 
pages of the book. It will be found, therefore, that while the size of the book 
is unusually large it will be less difficult to study than if it were short. 

As a text the book may be used for courses of practically any length, but 
it is not int ended. Vhat in any course on the subject every page of the book shall 
be used as assipued text. In the new graduate course in mechanical engineering 
at Columbia University, about three-fourths of the subject matter of the book 
will be so used for a course of about one hundred and twenty periods of one 
hour each. All of the book matter not specifically assigned as text or reference 
in a course on engineering thermodynamics in any school may profitably be 
taken up in courses on other subjects, serving more or less as a basis for them. 
It is therefore adapted to courses on gas power, compressed air, steam turbines, 
steam power plants, steam engine design, mechanical refrigeration, heating 
and ventilation, chemical factory equipment, laboratory practice and research. 
Whenever a short course devoted to engineering thermodynamics alone is 
desired, the earlier sections of each chapter combined in some cases with the 
closing sections, may be assigned as text. In this manner a course of about 
thirty hours may be profitably pursued. This is a far better procedure than 
using a short text to fit a short course, as the student gets a better perspective, 
and may later return to omitted topics without difficulty. 

The preparation of the manuscript involves such a great amount of labor, 
that it would never have been undertaken without the assurance of assistance 
by Mr. E. D. Thurston, Jr., a fellow instructor at Columbia, in checking text 
and tables, calculating diagrams, writing problems and working examples. 
This help has been invaluable and is gratefully acknowledged. Recognition 
is also due for material aid rendered by Mr. T. M. Gunn in checking and in 
some cases deriving formulas, more especially those of the first three chapters. 
In spite, however, of all care to avoid errors it is too much to expect complete 
success in a new work of this character, but it is hoped that readers finding 
errors will point them out that future editions may be corrected. 

C. E. L. 
COLUMBIA UNIVERSITY, NEW YORK, September, 1912, 



CONTENTS 



CHAPTER I. WORK AND POWER. GENERAL PRINCIPLES 

PAGE 

1. Work defined 1 

2. Power denned 2 

3. Work in terms of pressure and volume 3 

4. Work of acceleration and resultant velocity 6 

5. Graphical representation of work 8 

6. Work by pressure volume change 10 

7. Work of expansion and compression . 13 

8. Values of exponent s defining special cases of expansion or compression 20 

9. Work phases and cycles, positive, negative and net work 24 

10. Work determination by mean effective pressure .' 31 

11. Relation of pressure-volume diagrams to indicator cards 34 

12. To find the clearance 37 

.3. Measurement of areas of PV diagrams and indicator cards 43 

.4. Indicated horse-power 44 

.5. Effective horse-power, brake horse-power, friction horse-power, mechanical efficiency, 

efficiency of transmission, thermal efficiency : 47 

.6. Specific displacement, quantity of fluid per hour, or per minute per I.H.P 49 

.7. Velocity due to free expansion by PV method 52 

8. Weight of flow through nozzles by PV method 55 

9. Horse-power of nozzles and jets, by PV method 57 

CHAPTER II. WORK OF COMPRESSORS. HORSE-POWER and CAPACITY OF AIR, GAS AND 
VAPOR COMPRESSORS, BLOWING ENGINES AND DRY VACUUM PUMPS 

1. General description of structures and processes 73 

2. Standard reference diagrams or PV cycles for compressors and methods of analysis 

of compressor work and capacity 75 

3. Single-stage compressor, no clearance, isothermal compression, Cycle I. Work, 

capacity, and work per cubic foot in terms of pressures and volumes 81 

4. Single-stage compressor with clearance, isothermal compression, Cycle II. Work, 

capacity, and work per cubic foot in terms of pressures and volumes 85 

5. Single-stage compressor, isothermal compression. Capacity, volumetric efficiency, 

work, mean effective pressure, hor&e-power and horse-power per cubic foot of 

substance, in terms of dimensions and cylinder clearance 87 

6. Single-stage compressor, no clearance, exponential compression, Cycle III. Work, 

capacity and work per cubic foot, in terms of pressures and volumes 91 

7. Single-stage compressor with clearance, exponential compression, Cycle IV. Work, 

capacity, and work per cubic foot in terms of pressures and volumes 96 

8. Single-stage compressor, exponential compression. Relation between capacity, 

volumetric efficiency, work, mean effective pressure, horse-power and horse- 
power per cubic foot of substance and the dimensions of cylinder and clearance . . 98 

ix 



x CONTENTS 

PAGE 

9. Two-stage compressor, no clearance, perfect intercooling, exponential compression, 
best-receiver pressure, equality of stages. Cvcle V. Work and capacity in terms 
of pressures and volumes 103 

10. Two-stage compressor, with clearance, perfect intercooling 6x7 < nitial compression, 

best-receiver pressure, equality of stages, Cycle VI. Work and capacity in terms 

of pressures and volumes 109 

11. Two-stage compressor, any receiver pressure, exponential compression. Capacity, 

volumetric efficiency, work, mean effective pressure and horse-power, in terms of 
dimensions of cylinders and clearances 113 

12. Two-stage compressor, with best-receiver pressure, exponential compression. Capacity, 
volumetric efficiency, work, mean effective pressure and horse-power in terms 

of dimensions of cylinders and clearances 120 

13. Three-stage compressor, no clearance, perfect intercooling exponential compres- 

sion, best two receiver pressures, equality of stages, Cycle VII. Work and 
capacity, in terms of pressures and volumes 125 

14. Three-stage compressor with clearance, perfect intercooling exponential compression, 

best-receiver pressures, equality of stages, Cycle VIII. Work and capacity in 
terms of pressures and volumes 131 

15. Three-stage compressor, any receiver pressure, exponential compression. Capacity, 

volumetric efficiency, work, mean effective pressure, and horse-power in terms 

of dimensions of cylinders and clearances 135 

16. Three-stage compressor urith best-receiver pressures, exponential compression. Capac- 

ity, volumetric efficiency, work, mean effective pressure and horse-power in terms 

of dimensions of cylinders and clearances 143 

17. Comparative economy or efficiency of compressors 148 

18. Conditions of maximum work of compressors 151 

19. Compressor characteristics 153 

20. Work at partial capacity in compressors of variable capacity 160 

21. Graphic solution of compressor problems 168 

CHAPTER III. WORK OF PISTON ENGINES. HORSE-POWER AND CONSUMPTION OF 
PISTON ENGINES USING STEAM, COMPRESSED AIR, OR ANY OTHER GAS OR VAPOR 
UNDER PRESSURE 

1. Action of fluid in single cylinders. General description of structure and processes. . 187 

2. Simple engines. Standard reference cycles or PV diagrams for the work of expansive 

fluids in a single cylinder 192 

3. Work of expansive fluid in single cylinder without clearance. Logarithmic expan- 

sion, Cycle I. Mean effective pressure, horse-power and consumption of simple 
engines 197 

4. Work of expansive fluid in single cylinder without clearance. Exponential expan- 

sion, Cycle II. Mean effective pressure, horse-power and consumption of simple 
engines 205 

5. Work of expansive fluid in single cylinder with clearance. Logarithmic expansion 

and compression; Cycle III. Mean effective pressure, horse-power, and con- 
sumption of simple engines 208 

6. Work of expansive fluid in single cylinder with clearance; exponential expansion 

and compression, Cycle IV. Mean effective pressure, horse-power and consumption 

of simple engines 219 

7. Action of fluid in multiple-expansion cylinders. General description of structure 

and processes 225 

8. Standard reference cycles or PV diagrams for the work of expansive fluids in two- 

cylinder compound engines 235 



CONTENTS xi 

PAGE 

9. Compound engine with infinite receiver. Logarithmic law. No clearance, Cycle 

V. General relations between pressures, dimensions, and work 256 

0. Compound engine with infinite receiver. Exponential law. No clearance, Cycle 

VI. General relations between pressures, dimensions and work 268 

1. Compound engine with finite receiver. Logarithmic law. No clearance, Cycle. 

VII. General relations between dimensions and work when H.P. exhaust and 
L.P. admission are not coincident. . . 274 

2. Compound engine with finite receiver. Exponential law, no clearance, Cycle VIII. 

General relations between pressures, dimensions, and work, when high pressure 
Exhaust and low-pressure admission are independent 287 

3. Compound engine without receiver. Logarithmic law. No clearance, Cycle IX. 

General relations between dimensions and work when high-pressure exhaust 
and low-pressure admission are coincident 292 

4. Compound engine without receiver. Exponential law. No clearance, Cycle X. 

General relations between dimensions and work when high-pressure exhaust 
and low-pressure admission are coincident 301 

5. Compound engine with infinite receiver. Logarithmic law, with clearance and 

compression, Cycle XI. General relations between pressures, dimensions and 
work 306 

6. Compound engine with infinite receiver. Exponential law, with clearance and 

compression, Cycle XII. General relations between pressures, dimensions and 
work 319 

7. Compound engine with finite receiver. Logarithmic law, with clearance and com- 

pression, Cycle XIII. General relations between pressures, dimensions, and 
work when H.P. exhaust and L.P. admission are independent 325 

8. Compound engine with finite receiver. Exponential law, with clearance and com- 

pression, Cycle XIV. General relations between pressures, dimensions, and 
work when H.P. exhaust and L.P. admission are independent 335 

9. Compound engine without receiver. Logarithmic law, with clearance and com- 

pression, Cycle XV. General relations between pressures, dimensions, and 
work when H.P. exhaust and L.P. admission are coincident 339 

0. Compound engine without receiver. Exponential law, with clearance and compres- 

sion, Cycle XVI. General relations between pressures, dimension, and work, 

when H.P. exhaust and L.P. admission are coincident , 346 

1 1. Triple-expansion engine with infinite receiver. Logarithmic law. No clearance, 

Cycle XVII. General relations between pressures, dimensions and work 349 

2. Multiple-expansion engine. General case. Any relation between cylinders and 

receiver. Determination of pressure-volume diagram and work, by graphic 
methods - 357 

3. Mean effective pressure, engine horse-power, and work distribution and their vari- 

ation with valve movement and initial pressure. Diagram distortion and diagram 
factors. Mechanical efficiency 363 

1. Consumption of steam engines and its variation with valve movement and initial 

pressure. Best cut-off as affected by condensation and leakage 371 

5. Variation of steam consumption with engine load. The Willans line. Most eco- 
nomical load for more than one engine and best load division 381 

3. Graphical solution of problems on engine horse-power and cylinder sizes 387 



xii CONTENTS 



CHAPTER IV. HEAT AND MATTER. QUALITATIVE AND QUANTITATIVE RELATIONS 
BETWEEN HEAT CONTENT OF SUBSTANCES AND PHYSICAL-CHEMICAL STATE 

PAGE 

1. Substances and heat effects important in engineering. . ( 398 

2. Classification of heating processes. Heat addition and abstraction with or without 

temperature change, qualitative relations 401 

3. Thermometry based on temperature change, heat effects. Thermometer and abso- 

lute temperature scales 407 

4. Calorimetry based on proportionality of heat effects to heat quantity. Units of 

heat and mechanical equivalent 415 

5. Temperature change relation to amount of heat for solids, liquids, gases and vapors 

not changing state. Specific heats 419 

6. Volume or density variation with temperature of solids, liquids, gases and vapors 

not changing state. Coefficients of expansion. Coefficients of pressure change 
for gases and vapors 435 

7. Pressure, volume, and temperature relations for gases. Perfect and real gases 438 

8. Gas density and specific volume and its relation to molecular weight and gas constant . 446 

9. Pressure and temperature relations for vapor of liquids or solids. Vaporization, 

sublimation and fusion curves. Boiling- and freezing-points for pure liquids 
and dilute solutions. Saturated and superheated vapors 451 

10. Change of state with amount of heat at constant temperature. Latent heats of 

fusion and vaporization. Total heats of vapors. Relation of specific volume 

of liquid and of vapor to the latent heat 467 

11. Gas and vapor mixtures. Partial and total gas and vapor pressures. Volume, 

weight and gas constant relations. Saturated mixtures. Humidity 481 

12. Absorption of gases by liquids and by solids. Relative volumes and weights with 

pressure and temperature. Heats of absorption and of dilution. Properties 

of aqua ammonia 493 

13. Combustion and related reactions. Relative weights and volumes of substances 

and elements before and after reaction 506 

14. Heats of reaction. Calorific power of combustible elements and of simple chemical 

compounds. B.T.U. per pound and per cubic foot 516 

15. Heat transmission processes. Factors of internal conduction, surface resistance, 

radiation and convection 528 

16. Heat transmission between separated fluids. Mean temperature differences, coeffi- 

cients of transmission 538 

17. Variation in coefficient of heat transmission due to kind of substance, character of 

separating wall and conditions of flow v 555 

CHAPTER V. HEATING BY COMBUSTION. FUELS, FURNACES, GAS PRODUCERS AND 
STEAM BOILERS 

1. Origin of heat and transformation to useful form. Complexity of fuels as sources 

of heat. General classification, solid, liquid, gaseous, natural and artificial 644 

2. Natural solid fuels, wood peat, lignite, bituminous and anthracite coals. Chemical 

and physical properties. Classifications based on ultimate and proximate analysis 
and on behavior on heating 649 

3. Calorific power of coals and the combustible of coals. Calculation of calorific power 

from ultimate and proximate analyses. Calorific power of the volatile 662 

4. Mineral oil and natural gas fuel. Chemical and physical properties. Calorific 

power direct and as calculated for oils from ultimate analysis or from density, 
and for gas from sum of constituent gases 67C 






CONTENTS xiii 



PAGE 



5. Charcoal, coke, coke oven and retort coal gas as products of heating wood and coal. 

Chemical, physical, and calorific properties per pound. Calorific power of gases 
per cubic foot in terms of constituent gases. Yield of gas and coke per pound 
coal 675 

6. Distillate oils, kerosene and gasoline, residue oils, and oil gas as products of heating 

mineral oils. Chemical, physical, and calorific properties. Calorific power of frac- 
tionated oils in terms of, (a) carbon and hydrogen; (6) density per pound, and 
estimated value per cubic foot vapor. Calorific power of oil gas per pound and per 
cubic foot in terms of constituent gases. Yield of distillates and oil gas 685 

7. Gasification of fixed carbon and coke by air-blast reactions, producing ah* gas, and 

blast-furnace gas. Comparative yield per pound coke and air. Sensible heat 
and heat of combustion of gas. Relation of constituents in gas. Efficiency 
of gasification 695 

8. Gasification of fixed carbon, coke, and coal previously heated, by steam-blast reac- 

tions, producing water gas. Composition and relation of constituents of water 
gas, yield per pound of steam and coal. Heat of combustion of gas and limitation 
of yield by negative heat of reaction 710 

9. Gasification of coals by steam and air blasts, resulting in producer gas. Composition 

and relation of constituents of producer gas, yield per pound of fixed carbon, air 
and steam. Modification of composition by addition of volatile of coal. Heat 
of combustion of gas, sensible heat, and efficiency of gasification. Horse-power 
of gas producers 719 

10. Combustion effects. Final temperature, volume and pressure for explosive and 

non-explosive combustion. Estimation of air weights and heat suppression 
due to CO in products from volumetric analysis 740 

11. Temperature of ignition and its variation with conditions. Limits of proportion 

air gas neutral, or detonating gas and neutral, for explosive combustion of mix- 
tures. Limits of adiabatic compression for self-ignition of mixtures 758 

12. Rate of combustion of solid fuels with draft. Propagation rates, uniform and 

detonating for explosive gaseous mixtures 765 

13. Steam-boiler evaporative capacity and horse-power. Horse-power units, equivalent 

rates of evaporation and of heat absorption. Factors of evaporation. Relation 
between absorption rates and rates of heat generation. Influence of heating and 
grate surface, calorific power of fuels, rates of combustion and furnace losses. . . . 773 

14. Steam-boiler efficiency, furnace and heating-surface efficiency. Heat balances and 

variation in heat distribution. Evaporation and losses per jpound of fuel 796 

CHAPTER VI. HEAT AND WORK. GENERAL RELATIONS BETWEEN HEAT AND WORK. 
THERMAL EFFICIENCY OF STEAM, GAS, AND COMPRESSED-AIR ENGINES. FLOW 
OF EXPANSIVE FLUIDS. PERFORMANCE OF MECHANICAL REFRIGERATING SYSTEMS 

1. General heat and work relations. Thermal cycles. Work and efficiency deter- 

mination by heat differences and ratios. Graphic method of temperature 
entropy heat diagram , 874 

2. General energy equation between heat change, intrinsic energy change, and work 

done. Derived relations between physical constants for gases and for changes 

of state, solid to liquid, and liquid to vapor 882 

3. Quantitative relations for primary thermal phases, algebraic, and graphic to PV, 

and T3> coordinates. Constancy of PV, and T for gases and vapors, wet, dry 
and superheated 892 

4. Quantitative relations for secondary thermal phases. Adiabatics for gases and 

vapors. Constant quality, constant total heat, and logarithmic expansion lines 

for steam. . 904 



xiv CONTENTS 



5. Thermal cycles representative of heat-engine processes. Cyclic efficiency. A 

reference standard for engines and fuel-burning power systems. Classification 
of steam cycles 

6. The Rankine cycle. Work, mean effective pressure, jet velocity, water rate, heat 

consumption and efficiency of steam Cycle I. Adiabatic expansion, constant 
pressure, heat addition and abstraction, no compression 936 

7. Carnot cycle and derivatives. Work, mean effective pressure, water rate, heat 

consumption and efficiency of steam Cycle II. Adiabatic expansion and com- 
pression, constant pressure heat addition and abstraction 957 

8. Gas cycles representative of ideal processes and standards of reference for gas 

engines 970 

9. Brown, Lenoir, Otto and Langen non-compression gas-cycles. Work, mean 

effective pressure, volume and pressure ranges, efficiency, heat and gas con- 
sumption 978 

10. Stirling and Ericsson gas cycles. Work, efficiency and derived quantities for 

isothermal compression with and without regenerators 993 

11. Otto, Atkinson, Brayton, Diesel, and Carnot gas cycles. Work, efficiency and 

derived quantities for adiabatic compression gas cycles 1006 

12. Comparison of steam and gas cycles with the Rankine as standard for steam, and 

with the Otto and Diesel, as standard for gas. Relations of Otto and Diesel 

to Rankine cycle. Conditions for equal efficiency 1031 

13. Gas cycle performance as affected by variability of the specific heats of gases, 

applied to the Otto cycle 1035 

14. Actual performance of Otto and Diesel gas engines, and its relation to the cyclic. 

Diagram factors for mean effective pressure and thermal efficiency. Effect 

of load on efficiency. Heat balance of gas engines alone and with gas producers 1042 

15. Actual performance of piston steam engines and steam turbines at their best load 

and its relation to the cyclic. Effect on efficiency of initial pressure, vacuum, 
superheat, jacketing, and reheating. Heat balances of steam power plants .... 1062 

16. Flow of not water, steam and gases through orifices and nozzles. Velocity, weight 

per second, kinetic energy, and force of reaction of jets. Nozzle friction and 
reheating and coefficient of efflux. Relative proportions of series nozzles for 
turbines for proper division of work of expansion 1083 

17. Flow of expansive fluids under small pressure drops through orifices, valves, and 

Venturi tubes. Relation between loss of pressure and flow. Velocity heads 
and quantity of flqw by Pitot tubes 1097 

18. Flow of gases and vapors in pipes, flues, ducts, and chimneys. Relation between 

quantity of flow and loss of pressure. Friction resistances. Draught and 
capacity of chimneys 1111 

19. Thermal efficiency of compressed-air engines alone and in combination with air 

compressors. Effect of preheating and reheating. Compressor suction heating, 

and volumetric efficiency. Wall action 1127 

20. Mechanical refrigeration, general description of processes and structures. Thermal 

cycles and refrigerating fluids. Limiting temperatures and pressures 1142 

21. Performance of mechanical refrigerating cycles and systems. Quantity of fluid 

circulated per minute per ton refrigeration, horse-power, and heat supplied 
per ton. Refrigeration per unit of work done and its relation to thermal effi- 
ciency of the system 1157 



LIST OF TABLES 



NO. PAGE 

1. Conversion table of units of distance 62 

2. Conversion table of units of surface 62 

3. Conversion table of units of volume 62 

4. Conversion table of units of weights and force 63 

5. Conversion table of units of pressure 63 

6. Conversion table of units of work 64 

7. Conversion table of units of power 64 

8. Units of velocity 64 

9. Barometric heights, altitudes and pressures 65 

10. Values of s in the equation PV S = constant for various substances and conditions . . 67 

11. Horse-power per pound mean effective pressure 68 

12. Ratio of cut-offs in the two cylinders of the compound engine to give equal work 

for any receiver volume 284 

13. Piston positions for any crank angle 395 

14. Values for x for use in Heck's formula for missing water 396 

15. Some actual steam engine dimensions 396 

16. Fixed temperatures 411 

17. Fahrenheit temperatures by hydrogen and mercury thermometers 414 

18. Freezing-point of calcium chloride brine ; 425 

19. Specific heat of sodium chloride brine 427 

20. Specific heat and gas constants, 431 

21. The critical point 453 

22. Juhlin's data on the vapor pressure of ice 456 

23. Tamman's value on fusion pressure and temperature of water-ice 456 

24. Lowering of freezing-points 465 

25. Berthelot's data on heat for complete dilution of ammonia solutions 500 

26. Air required for combustion of various substances 515 

27. Radiation coefficients 535 

28. Coefficients of heat transfer 550 

29. Temperatures, Centigrade and Fahrenheit 571 

30. Heat and power conversion table 573 

31. Specific heat of solids 574 

32. Specific heats of liquids 576 

33. Baume-specific gravity scale 577 

34. Specific heats of gases 578 

35. Coefficient of linear expansion of solids 580 

36. Coefficient of cubical expansion of solids 581 

37. Coefficient of volumetric expansion of gases and vapors at constant pressure 582 

xv 



xv i LIST OF TABLES 

NO, PAGE 

38. Coefficient of pressure rise of gases and vapors at constant volume 583 

39. Compressibility of gases by their isothermals 584 

40. Values of the gas constant R 584 

41. Density of gases 585 

42. International atomic weights 586 

43. Melting- or freezing-points 586 

44. Boiling-points 588 

45. Latent heats of vaporization 590 

46. Latent heats of fusion 591 

47. Properties of saturated steam 592 

48. Properties of superheated steam 596 

49. Properties of saturated ammonia vapor 603 

50. Properties of saturated carbon dioxide vapor 618 

51. Relation between pressure, temperature and per cent NH 3 in solution 628 

52. Values of partial pressure of ammonia and water vapors for various temperatures 

and per cents of ammonia in solution 632 

53. Absorption of gases by liquids 634 

54. Absorption of air in water 635 

55. Heats of combustion of fuel elements and chemical compounds 636 

56. Internal thermal conductivity 639 

57. Relative thermal conductivity 642 

58. General classification of fuels 648 

59. Comparison of cellulose and average wood compositions 650 

60. Classification of coals by composition 652 

61. Classification of coals by gas and coke qualities 654 

62. Composition of peats 655 

63. Composition of Austrian lignites 656 

64. Composition of English coking coals 658 

65. Wilkesbarre anthracite coal sizes and average ash content 659 

66. Density and calorific power of natural gas 673 

67. Products of wood distillation 676 

68. Products of peat distillation 678 

69. Products of bituminous coal distillation 680 

70. Gas yield of English cannel coals 682 

71. Comparison of coke oven and retort coal gas 682 

72. Relation between oxygen in coal and hydrocarbon in gas 684 

73. Density and calorific power of coke oven gas 684 

74. Average distillation products of crude mineral oils 686 

75. American mineral oil products 687 

76. U.S. gasolene and kerosene bearing crude oils 688 

77. Calorific power of gasolenes and kerosenes 691 

78. Properties of oil-gas 693 

79. Yield of retort oil gas 694 

80. Density and calorific power of oil gas 694 

81. Boudouard's equilibrium relations for CO and CO2 with temperature 697 

82. Change of O 2 in air to CO and CO 2 at 1472 F 699 

83. Composition of hypothetical air gas, general 704 

84. Composition of hypothetical air gas, no CO 2 and no CO 705 

85. Density and calorific power of blast furnace gas 708 

86. Water gas characteristics with bed temperature 710 

87. Composition of hypothetical water gas, general 714 

88. Composition of hypothetical water gas, no CO 2 and no CO . 715 



LIST OF TABLES xvii 

*O. PAGE 

89. Density and calorific power of water gas 718 

90. Composition of hypothetical producer gas from fixed carbon 725 

91. Density and calorific power of producer gas 737 

92. Characteristics of explosive mixtures of oil gas and air 747 

93. Calculated ignition temperatures for producer gas 761 

94. Compressions commonly used in gas engines 762 

95. Ignition temperatures 763 

96. Variation of ignition temperature of charcoal with distillation temperature 763 

97. Per cent detonating mixture at explosive limits of proportion 764 

98. Velocity of detonating or explosive waves 766 

99. Rate of propagation, feet per second, for different proportions in gaseous mixtures . 768 

.00. Rates of combustion for coal 769 

lOl. Constants of proportion for rate of coal combustion for use in Eq. (848) 771 

.02. Boiler efficiency summaries 799 

103. Three examples of heat balance for boilers 800 

'.04. Composition and calorific power of characteristic coals 818 

.05. Combustible and volatile of coals, lignites and peats 826 

i. Paraffines from Pennsylvania petroleums 835 

i.07. Calorific power of mineral oils by calorimeter and calculation by density formula 

of Sherman and Kropff 836 

108. Properties of mineral oils 838 

109. Composition of natural gases 841 

110. Composition of coke oven and retort coal gas 842 

111. Composition of U. S. coke 846 

112. Fractionation tests of kerosenes and petroleums 847 

113. Fractionation tests of gasolenes 851 

114. Composition of blast-furnace gas and air gas 853 

LI 5. Rate of formation of CO from CO 2 and carbon 855 

L16. Composition of water gas 857 

117. Composition of producer gas 858 

118. Gas producer tests 864 

119. Composition of oil producer gas 866 

120. Composition of powdered coal producer gas 866 

121. Calorific powers of best air-gas mixtures 867 

122. Composition of boiler-flue gases 868 

123. Limits of proportions of explosive air-gas mixtures 869 

!124. Rate of combustion of coal with draft 870 
125. Rate of combustion of coal 871 

|126. Values of s for adiabatic expansion of steam 912 

1 127. Values of s for adiabatic expansion of steam determined from initial and final volumes 

only 913 

I.L28 1042 

129. Diagram factors for Otto cycle gas engines 1046 

130. Mechanical efficiencies of gas engines 1050 

: 131. Allowable compression for gas engines 1050 

1 132. Mean effective pressure factors for Otto cycle engines 1053 

jL33. Giildner's values of Otto engine real volumetric efficiency with estimated mean 

suction resistances 1055 

1 134. Comparative heat balances of gas producer and engine plants 1057 

j 135. Heat balances of gas producer plants 1060 

JL36. Heat balances of gas and oil engines 1060 

137. Steam plant heat balances 1063 



xviii LIST OF TABLES 

NO. PA< 

138. Efficiency factors for reciprocating steam engines and turbines 10( 

139. Steam turbine efficiency and efficiency factors with varying vacuum and with 

steam approximately at constant initial pressure 10' 

140. Efficiency factors for low-pressure steam in piston engines 10' 

141. Coefficient of discharge for various air pressure and diameters of orifice (Durley) . IK 

142. Values of C for air flow (Weisbach) Ill 

143. Flow change resistance factors FR (Reitschel) 11' 



TABLE OF SYMBOLS 



A =area in square feet. 

= constant, in formula for most economical load of a steam engine, Chapter III. 

= constant, in pipe flow formula^ Chapter VI. 

= excess air per pound of coal, Chapter V. 

= pounds of ammonia dissolved per pound of weak liquor, Chapter IV. 
a = area in square inches. 

= coefficient of linear expansion, Chapter IV. 

= constant in equation for the ratio of cylinder sizes for equal work distribution in com- 
pound engine, Chapter III. 

= constant in equation for change in intrinsic energy, Chapter VI. 

= constant in equation for specific heat, Chapter IV. 

= cubic feet of air per cubic foot of gas in explosive mixtures, Chapter' V. 

= effective area of piston, square inches, Chapter I. 



B = constant in equation for the most economical load of the steam engine, Chapter III. 

= constant in equation for flow in pipes, Chapter VI. 
Be. = Baume. 
B.H.P. = brake horse-power, Chapters III and VI. 

= boiler horse-power, Chapter V. 
B.T.U. = British thermal unit. 

6 = constant in equation for change in intrinsic energy, Chapter VI. 

= constant in equation for specific heat, Chapter IV. 
(bk.pr.) =back pressure in pounds per square inch. 



C = Centigrade. 

= circumference or perimeter of ducts in equations for flow, Chapter VI. 
= constant. 

= heat suppression factor, Chapter V. 
= ratio of pressure after compression to that before compression in gas engine cycles, 

Chapter VI. 

= specific heat, Chapter IV. 

Cc per cent of ammonia in weak liquor, Chapter VI. 
C p = specific heat at constant pressure. 
C/z=per cent of ammonia in rich liquor, Chapter VI. 
C s = specific heat of water, Chapter VI. 
Cv specific heat at constant volume. 
Cl = clearance expressed in cubic feet. 
c = clearance expressed as a fraction of the displacement 

= constant, 
cu.ft. = cubic foot, 
cu. in. = cubic inch. 



D = constant in equations for pipe flow, Chapter VI. 

= density, Chapter IV. 

= diameter of pipe in feet, Chapter VI. 

= displacement in cubic feet. 
D s = specific displacement, Chapter I. 

xix 



xx TABLE OF SYMBOLS 

d = constant in equation for change in intrinsic energy, Chapter VI. 
= diameter of a cylinder in inches, Chapter I. 
= diameter of pipe in inches, Chapter VI. 
= differential, 
(del.pr.) = delivery pressure in pounds per square inch, Chapter II. 



E = constant in equation for pipe flow, Chapter VI. 
= external latent heat, Chapter IV. 
= thermal efficiency, Chapter VI. 

EB = thermal efficiency referred to brake horse-power, Chapter III. 
Efy = boiler efficiency, Chapter V. 
Ef = furnace efficiency, Chapter V. 

EI = thermal efficiency referred to indicated horse-power, Chapter III. 
E m = mechanical efficiency, Chapter III. 
E s = heating surface efficiency, Chapter V. 
EV= volumetric efficiency (apparent), Chapter VI. 
EV' = volumetric efficiency (true), Chapter VI. 
e = as a subscript to log to designate base e. 

= constant in equation for change in intrinsic energy, Chapter VI. 
ei= ratio of true volumetric efficiency to hypothetical, Chapter II. 
62 = ratio of true volumetric efficiency to apparent, Chapter III. 
e z = ratio of true indicated horse-power to hypothetical, Chapter II. 



F = constant in equation for pipe flow, Chapter VI. 
= diagram factor for gas engine indicator cards, Chapter VI. 
= Fahrenheit. 
= force in pounds. 

F F = friction factor, F/?Xvelocity head = loss due to friction, Chapter VI. 
FR = resistance factor, FRX velocity head=loss due to resistances, Chapter VI. 
Fs=special resistances to flow in equations for chimney draft, Chapter VI. 
/= constant in equation for changes in intrinsic energy, Chapter VI. 

= function. 
ft.=foot. 
ft.-lb. = foot-pound. 



G = constant in equation for pipe flow, Chapter VI. 

= weight of gas per hour in equation for chimney flow, Chapter VI. 
G m = maximum weight of gases in equation for chimney flow, Chapter VI. 
G. S. = grate surface. 

g = acceleration due to gravity, 32.2 (approx.) feet per second, per second. 



H = as a subscript to denote high pressure cylinder. 
= heat per pound of dry saturated vapor above 32 F. 
= heat per cubic foot gas. 
= heat transmitted, Chapter IV. 
= height of column of hot gases in feet, Chapter VI. 
= pressure or head in feet of fluid, Chapter VI. 

HA = difference in pressure on two sides of an orifice in feet of air, Chapter VI. 
Ho = equivalent head of hot gases, Chapter VI. 
H M = pressure in feet of mercury, Chapter VI. 
H w = pressure in feet of water, Chapter VI. 
H.P. = high pressure. 

= horse-power, Chapter I. 



TABLE OF SYMBOLS xxi 

H.S. = heating surface. 
H.P.cap.)=bigh pressure cylinder capacity, Chapter III. 

h = heat of superheat. 

Ji M = difference in pressure on two sides of an orifice in inches of mercury, Chapter VI. 
hw = difference in pressure on two sides of an orifice in inches of water, Chapter VI. 



7 = as a subscript to denote intermediate cylinder, Chapter III. 
I.H.P. = indicated horse-power. 

in. = inch, 
in.pr.) = initial pressure in pounds per square inch. 



7= Joule's equivalent = 778 (approx.) foot-pounds per B.T.U. 



K = coefficient of thermal conductivity, Chapter IV. 
= constant. 
= proportionality coefficient in equation for draft, Chapter VI. 

K e = engine constant =~~ in expression for horse-power. Chapter III. 
oo()00 



L=as a subscript to denote low-pressure cylinder. 
= distance in feet. 

= latent heat, Chapters IV and VI. 
= length of stroke in feet, Chapter I. 
L=per cent of heat in fuel lost in furnace, Chapter V. 
L.P. =low pressure. 
(L. P. Cap. )= low-pressure capacity, Chapter II. 
1 = constant, Chapter III. 

= length, Chapter IV. 
Ib. = pound. 

log = logarithm to the base 10. 
log e = logarithm to the base e. 



M = mass. 

(M.E.P.) =mean effective pressure, pounds per square foot. 
m = constant, Chapter III. 

area 

= mean hydraulic radius = . 
perimeter 

= molecular weight, Chapter IV. 

= ratio of initial pressure to that end of expansion in Otto and Langen gas cycle, 

Chapter VI. 

(m.b.p.) =mean back pressure in pounds per square inch, 
(m.e.p.) =mean effective pressure in pounds per square inch, 
(m.f.p.) =mean forward pressure in pounds per square inch. 



N = constant, Chapter III. 

= revolutions per minute. 
n = cycles per minute. 

= constant, Chapter III. 

= cubic foot of neutral per cubic foot of gaseous mixture, Chapter V. 

= number of degrees exposed on thermometer stem, Chapter IV. 

= ratio of volume after expansion to volume before in Atkinson gas cycle, Chapter IV. 

= specific volume of dry saturated steam, Chapter VI. 



xxii TABLE OF SYMBOLS 

O= volume of receiver of compound engine in cubic feet, Chapter III. 
P = draft in pounds per square foot, Chapter VI. 
= load in kilowatts, Chapter III. 
= pressure in pounds per square foot. 

PF = static pressure in pounds per square foot lost in wall friction, Chapter VI. 
PR= static pressure in pounds per square foot lost in changes of cross-section, etc., 

Chapter VI. 

PV= velocity head in pounds per square foot. 
p= pressure in pounds per square inch. 
p c = mean exhaust pressure, Chapter VI. 
p s = mean suction pressure, Chapter VI. 
pv= partial pressure of water vapor in air, Chapter VI. 



Q = quantity of heat or energy in B.T.U. gained or lost by a body passing from one state to 

another. 

Q/=heat added from fire in Stirling and Ericsson cycles, Chapter VI. 
Qi"=heat added from regenerator in Stirling and Ericsson cycles, Chapter VI. 
Q2 /= heat abstracted by water jacket in Stirling and Ericsson cycles, Chapter VI. 
Q 2 "=heat abstracted by regenerator in Stirling and Ericsson cycles, Chapter VI. 

q = quantity of heat per pound of liquid above 32 F. 



R ratio of heating surface to grate surface, Chapter V. 
' =gas constant. 
* RC^ ratio of cylinder sizes in two-stage air compressor or compound engine, Chapters II 

and III. 

RH= ratio of expansion in high-pressure cylinder, Chapter III. 
RL= ratio of expansion in low-pressure cylinder, Chapter III. 
JRp=ratio of initial to back pressure, Chapters III and VI. 
R p = ratio of delivery to supply pressure, Chapter II. 
Rv = ratio of larger volume to smaller volume. 

r=rate of flame propagation in explosive mixtures, Chapter V. 
r/>=pressure differences (maximum minimum) in gas cycles, Chapter VI. 
rv= volume differences (maximum minimum) in gas cycles, Chapter VI. 
(rec.pr.) = receiver pressure in pounds per square inch, Chapter III. 
(rel.pr.) = release pressure in pounds per square inch, Chapter III. 



S =per cent of ammonia in solution, Chapter IV. 
= piston speed, Chapter I. 

= pounds of steam per pound of air in producer blast, Chapter V. 
= specific heat, Chapter IV. 

= specific heat of superheated steam, Chapter VI. 
(Sup.Vol.) = volume of steam supplied to the cylinder per stroke, Chapter III. 

s = general exponent of V in expansion or compression of gases. 
sp.gr. = specific gravity, 
sp.ht. = specific heat, 
sq.ft. = square foot, 
sq.in. = square inch, 
(sup.pr.) = supply pressure, in pounds per square inch. 



T = temperature, degrees absolute. 
TC = temperature of air, Chapter VI. 
TH = temperature of gases in chimney, Chapter VI. 
J= temperature in degrees scale. 



TABLE OF SYMBOLS xxiii 

7= rate of heat transfer in B.T.U. per square foot per hour per degree difference in tem- 
perature, Chapter IV. 
7 = intrinsic energy, Chapter VI. 
u = velocity in feet per second. 
m = velocity in feet per minute, Chapter VI. 



V= volume in cubic feet. 
7 A = cubic feet per pound air, Chapter VI. 
'a = cubic feet per pound, gas, Chapter VI. 
7 L = volume of liquid in cubic feet per pound. 
r s = volume of solid in cubic feet per pound. 
V = volume of vapor in cubic feet per pound. 
v= volume, Chapter IV. 



W = work in foot-pounds. 
V.R. = water rate. 

w = pounds of water per pound of ammonia in solution, Chapter IV. 

= weight in pounds. 
WR= pounds of rich liquor per pound of ammonia, Chapter VI. 



= compression in the steam engine as a fraction of the stroke, Chapter III. 

heat added 

= H , Chapter VI. 

temperature at beginning of addition X specific heat at constant volume 

x = constant in the expression for missing water, Chapter III. 
= fraction of liquid made from solid or vapor made from liquid, Chapter VI. 
= per cent of carbon burned to CO 2 , Chapter V. 
= per cent of nozzle reheat, Chapter VI. 
= per cent of steam remaining in high-pressure cylinder of compound engine at any point 

of the exhaust stroke, Chapter III. 
= quantity of heat added in generator of absorption system in addition to the amount of 

heat of absorption of 1 Ib. of ammonia, Chapter VI. 
ratio of low-pressure admission volume to high-pressure admission volume, Chapter III. 



Y = total steam used per hour by an engine, Chapter III. 

heat added 

= 1 + , Chapter VI. 

temperature at beginning of addition X specific heat at constant pressure 

j/ = per cent of vane reheat Chapter VI. 

= ratio of the volume of receiver to that of the high-pressure cylinder of the compound 
engine, Chapter III. 



Z= fraction of the stroke of the steam engine completed at cut-off, Chapter III. 

heat added from regenerator 

= 1+- , Chapter VI. 

temperature at beginning of addition X specific heat at constant volume 

'= hypothetical best value of Z. 

heat added from regenerator 

, Chapter VI. 

temperature at beginning of addition X specific heat at constant pressure 

2= ratio of R. P.M. to cycles per minute. 



ce = an angle, Chapter I. 

= coefficient of cubical expansion, Chapter III. 
= constant in the equation for latent heat, Chapter VI. 



xxiv TABLE OF SYMBOLS 

= constant in equation for variable specific heat at constant volume, Chapter VI. 
a' = constant in equation for variable specific heat at constant pressure, Chapter VI. 



= constant in equation for latent heat, Chapter VI. 

= fraction of fuel heat available for raising temperature, Chapter V. 



Y = constant in equation for latent heat, Chapter VI. 
= ratio of cross-section to perimeter, Chapter IV. 

sp. ht. at const, press. 

= special value for s for adiabatic expansion or compression = 

sp. ht. at const, vol. 

' = ratio of specific heat at constant pressure to specific heat at constant volume when 
is a variable, Chapter VI. 



A = increment. 

8 = density in pounds per cubic foot. 

,= density in cold gases in equations for chimney draft, Chapter VI. 
$H = density of hot gases in equations for chimney draft, Chapter VI. 



= coefficient of friction, Chapter VI. 



[i = material coefficient in heat transfer expression, Chapter IV. 



p= internal thermal resistance, Chapter IV. 



S= summation. 

a = surf ace thermal resistance, Chapter IV. 



u = time in seconds. 

$ = entropy, Chapter VI. 
<J> = entropy, Chapter VI. 



NOTE. A small letter when used as a subscript to a capital in general refers to a po 
on a diagram, e.g., P a designates pressure at the point A. Two small letters used as si 
scripts together, refer in general to a quantity between two points, e.g., W a b designa 
work done from point A to point B. 



ENGINEERING THERMODYNAMICS 



CHAPTER I 
WORK AND POWER. GENERAL PRINCIPLES. 

1. Work Defined. Work, in the popular sense of performance of any labor, 
is not a sufficiently precise term for use in computations, but the analytical 
mechanics has given a technical meaning to the word which is definite and which 
is adopted in all thermodynamic analysis. The mechanical definition of work is 
mathematical inasmuch as work is always a product of forces opposing motion 
and distance swept through, the force entering with the product being limited 
to that acting in the direction of the motion. The unit of distance in the 
English system is the foot, and of force the pound, so that the unit of work is 
the foot-pound. In the metric system the distance unit is the meter and the 
force unit the kilogramme, making the work unit the kilogrammeter. Thus, 
the lifting of one pound weight one foot requires the expenditure of one foot- 
pound of work, and the falling of one pound through one foot will perform one 
foot-pound of work. It is not only by lifting and falling weights that work is 
expended or done; for if any piece of mechanism be moved through a distance 
of one foot, whether in a straight or curved path, and its movement be resisted 
by a force of one pound, there will be performed one foot-pound of work against 
the resistance. It is frequently necessary to transform work from one sys- 
tem of units to the other, in which case the factors given at the end of this 
Chapter are useful. 

Work is used in the negative as well as in the positive sense, as the force 
considered resists or produces the motion, and there may be both positive and 
negative work done at the same time; similar distinctions may be drawn with 
reference to the place or location of the point of application of tne force. Con- 
sider, for example, the piston rod of a direct-acting pump in which a certain 
force acting on the steam end causes motion against some less or equal force 
acting at the water end. Then the work at the steam end of the pump may be 
considered to be positive and at the water end negative, so far as the move- 
ment of the rod is concerned; when, however, this same movement causes a 
movement of the water, work done at the water end (although negative with 
reference to the rod motion, since it opposes that motion) is positive with refer- 



2 ENGINEERING THERMODYNAMICS 

ence to the water, since it causes this motion. It may also be said that the 
steam does work on the steam end of the rod and the water end of the rod does 
work on the water, so that one end receives and the other delivers work, the rod | 
acting as a transmitter or that the work performed at the steam end is the input 
and that at the water end the output work. 

(See the end of Chapter I for Tables I, II, III, IV, and VI, Units of 
Distance, of Surface, of Volume, of Weight and Force, and of Work.) 

Example. An elevator weighing 2000 Ibs. is raised 80 ft. How much work is done in 
foot-pounds? 

Foot-pounds = force X distance 

=2000 X80 = 160,000 ft.-lbs. 

Ans. 160,000 ft.-lbs. 

Prob. 1. A pump lifts 150 gallons of water to a height of 250 ft. How much work does 
it do? i 

Prob. 2. By means of a jack a piece of machinery weighing 10 tons is raised f in. What 
is the work done? 

Prob. 3. A rifle bullet weighing 2 oz. travels vertically upward lj miles. What work was 
done in foot-pounds? 

Prob. 4. A cubic foot of water falls 50 ft. in reaching a water-wheel. How much work can 
it do? 

Prob. 5. A piston of an elevator is 12 ins. in diameter and has acting on it a pressure of 
80 Ibs. per square inch. What work is done per foot of travel? 

Prob. 6. It has been found that a horse can exert 75 Ibs. pull when going 7 miles per 
hour. How much work can be done per minute? 

Prob. 7. How much work is done by an engine which raises a 10-ton casting 50 ft.? 

Prob. 8. The pressure of the air on front of a train is 50 Ibs. per square foot when the 
speed is 50 miles per hour. If the train presents an area of 50 sq.ft., what work is done in 
overcoming wind resistance? 

Prob. 9. The pressure in a 10-inch gun during the time of firing is 2000 Ibs. per square 
inch. How much work is done in ejecting the projectile if the gun is 33 ft. 4 ins. long? 

2. Power Defined. Power is defined as the rate of working or the work 
done in a given time interval, thus introducing a third unit of mechanics, time, 
so that power will always be expressed as a quotient, the numerator being a prod- 
uct of force and distance, and the denominator time. This is in opposition 
to the popular use of the word, which is very hazy, but is most often applied to 
the capability of performing much work] or the exertion of great force, thus, 
popularly, a powerful man is one who is strong, but in the technical sense a man 
would be powerful only when he could do much work continuously and rapidly. 
An engine has large power when it can perform against resistance many foot- 
pounds per minute. The unit of power in the English system is the horse-power, 
or the performance of 550 foot-pounds per second or 33,000 foot-pounds per 
minute, or 1,980,000 foot-pounds per hour. In the metric system the horse- 
power is termed cheval-vapeur, and is the performance of 75 killogrammeters 
= 5422 foot-pounds per second, or 4500 kilogrammeters = 32,549 foot-pounds 



WORK AND POWER. 3 

3er minute, or 270,000 kilogrammeters = 1,952,932 foot-pounds per hour. 
Fable VII at the end of Chapter I gives conversion factors for power units. 

Example. The piston of a steam engine travels 600 ft. per minute and the mean force 
)f steam acting upon it is 65,000 Ibs. What is the horse-power? 

Horse-power = foot-pounds per minute 
oo,UUU 

^, distance 
force XT 
time 



33,000 
65,000X600 
~~ 



Prob. 1. The draw-bar pull of a locomotive is 3000 Ibs. when the train is traveling 50 
miles per hour. What horse-power is being developed? 

Prob. 2. A mine cage weighing 2 tons is lifted up a 2000-ft. shaft in 40 seconds. What 
horse-power will be required if the weight of the cable is neglected? 

Prob. 3. By direct pull on a cable it is found possible to lift 4 tons 20 ft. per second. With 
a differential pulley 40 tons may be lifted 3 ft. per second. What is the difference in power 
required? 

Prob. 4. A horse exerts a pull of 100 Ibs. on a load. How fast must the load be moved to 
develop one horse-power? 

Prob. 5. The resistance offered to a ship at a speed of 12 knots was 39,700 Ibs. What 
horse-power must be available to maintain this speed? (One knot is a speed of one nautical 
mile per hour.) 

Prob. 6. It is estimated that 100,000 cu. ft. of water go over a fall 60 ft. high every 
second. What horse-power is going to waste? 

Prob. 7. The force acting on a piston of a pump is 80,000 Ibs. If the piston speed is 150 
ft. per minute, what is the horse-power? 

Prob. 8. To draw a set of plows 2j miles per hour requires a draw-bar pull of 10,000 
Ibs. What must be the horse-power of a tractor, to accomplish this? 

Prob. 9. The horse-power to draw a car up a grade is the sum of the power necessary to 
pull it on a level and that necessary to lift it vertically the same number of feet as it rises on 
the grade. What will be the horse-power required to draw a car 20 miles per hour up a 12 per 
cent grade if the car weighs 2500 Ibs. and the draw-bar pull on the level is 250 Ibs.? 

3. Work in Terms of Pressure and Volume. Another of the definitions 
of mechanics fixes pressure as force per unit area so that pressure is always a 
quotient, the numerator being force and the denominator area, or length to 
the second power. If, therefore, the pressure of a fluid be known, and accord- 
ing to hydromechanics it acts equally and normally over all surface in contact 
with it, then the force acting in a given direction against any surface will be 
the product of the pressure and the projected area of the surface, the projection 
being on a plane at right angles to the direction considered. In the case of pis- 
tons and plungers the line of direction is the axis of the cylinder, and the pro- 
jected area is the area of the piston less the area of any rod passing completely 
through the fluid that may be so placed. When this plane area moves in a 



4 ENGINEEEING THERMODYNAMICS 

direction perpendicular to itself, the product of its area and the distance will be 
the volume swept through, and if a piston be involved the volume is technically 
the displacement of the piston. Accordingly, work may be expressed in three 
ways, as follows: 

Work = force X distance ; 

Work = pressure X area X distance ; 

Work = pressure X volume. 

The product should always be in foot-pounds, but will be, only when appro- 
priate units are chosen for the factors. These necessary factors are given as 
follows : 

Work in foot-pounds = force in Ibs. X distance in ft. 

= pressure in Ibs. per sq.ft. X area in sq.ft. X distance in ft 
= pressure in Ibs. per sq.in. X area in sq.in. X distance in ft 
= pressure in Ibs. per sq.ft. X volume in cu.ft. 
= pressure in Ibs. per sq.in. X 144 X volume in cu.ft. 

As pressures are in practice expressed in terms not only as above, but also 
in heights of columns of common fluids and in atmospheres, both in English and 
metric systems, it is convenient for calculation to set down factors of equivalence 
as in Table V, at the end of the Chapter. 

In thermodynamic computations the pressure volume product as an expres- 
sion for work is most useful, as the substances used are always vapors and gases, 
which, as will be explained later in more detail, have the valuable property of 
changing volume indefinitely with or without change of pressure according 
to the mode of treatment. Every such increase of volume gives, as a conse- 
quence, some work, since the pressure never reaches zero, so that to derive work 
from vapors and gases they are treated in such a way as will allow them to change 
volume considerably with as much pressure acting as possible. 

It should be noted that true pressures are always absolute, that is, measured 
above a perfect vacuum or counted from zero, while most pressure gages and 
other devices for measuring pressure, such as indicators, give results measured 
above or below atmospheric pressure, or as commonly stated, above or below 
atmosphere. In all problems involving work of gases and vapors, the absolute 
values of the pressures must be used; hence, if a gage or indicator measure- 
ment is being considered, the pressure of the atmosphere found by means of the 
barometer must be added to the pressure above atmosphere in order to obtain the 
absolute or true pressures. When the pressures are below atmosphere the 
combination with the barometric reading will depend on the record. If a record 
be taken by an indicator it will be in pounds per square inch below atmosphere 
and must be subtracted from the barometric equivalent in the same units to 
give the absolute pressure in pounds per square inch. When, however, a 
vacuum gage reads in inches of mercury below atmosphere, as such gages 
do, the difference between its reading and the barometric gives the absolute 



WORK AND POWER. 5 

-essure in inches of mercury directly, which can be converted to the desired 
aits 'by the proper factors. 

While it is true that the barometer is continually fluctuating at every place, 
frequently happens that standards for various altitudes enter into calculations, 
id to facilitate such work, values are given for the standard barometer at various 
titudes with equivalent pressures in pounds per square inch in Table IX. 

Frequently in practice, pressures are given without a definite statement 

: what units are used. Such a custom frequently leads to ambiguity, but it 

often possible to interpret them correctly from a knowledge of the nature of the 

roblem in hand. For instance, steam pressures stated by a man in ordinary prac- 

ce as being 100 Ibs. may mean 100 Ibs. per square inch gage (above atmosphere), 

ut may be 100 Ibs. per square inch absolute. Steam pressures are then most 

Dirimonly stated per square inch and should be designated as either gage or abso- 

ite. Pressures of compressed air are commonly expressed in the same units as 

team, either gage or absolute, though sometimes in atmospheres. Steam pressures 

elow atmosphere may be stated as a vacuum of so many inches of mercury, 

leaning that the pressure is less than atmosphere by that amount, or may 

e given as a pressure of so many inches of mercury absolute, or as so many 

ounds per square inch absolute. The pressures of gases stored in tanks under 

igh pressure are frequently recorded in atmospheres, due to the convenience 

I computation of quantities on this basis. Pressures of air obtained by blowers 

>r fans are usually given by the manufacturers of such apparatus in ounces 

>er square inch above (or below) atmosphere. Such pressures and also differ- 

nces of pressure of air due to chimney draft, or forced draft, and the pressure 

f illuminating gas in city mains, are commonly stated in inches of water, each 

nch of water being equivalent to 5.196 Ibs. per square foot. The pressure of 

vater in city mains or other pressure pipes may be stated either in pounds per 

iquare inch or in feet of water head. 

Example. A piston on which the mean pressure is 60 Ibs. per square inch sweeps through 
i volume of 300 cu.ft. What is the work done? 






W =PXV, where F=cu.ft. and P=lbs. per sq.ft. 
/. TF=60X144X300 =2,592,000 ft.-lbs. 



Prob. 1. The mean pressure acting per square inch when a mass of air changes in 
volume from 10 cu.ft. to 50 cu.ft. is 40 Ibs. per square inch. How much work is done? 

Prob. 2. An engine is required to develop 30 H.P. If the volume swept through per 
ninute is 150 cu.ft., what must the mean pressure be? 

Prob. 3. The mean effective pressure in compressing air from one to five atmospheres is 
28.7 Ibs. per square inch. How many horse-power are required to compress 1000 cu.ft. of 
free air per minute? 

Prob. 4. At an altitude of 4100 ft. a pressure gage showed the pressure on one side of 
l piston to be 50 Ibs. per square inch while the pressure on the opposite side is 3 Ibs. per 
square inch absolute. What pressure was tending to move the piston? 

Prob. 6. At an altitude of 1 mile the mean pressure in a gas engine cylinder during the 
suction stroke was found to be 12 Ibs. per square inch absolute. What work was done 
by the engine to draw in a charge if the cylinder was 5 ins. in diameter and the stroke 6 ins.? 



6 ENGINEERING THERMODYNAMICS 

Prob. 6. After explosion the piston of the above engine was forced out so that the ga 
volume was five times that at the beginning of the stroke. What must the M.E.P. hav<: 
been to get 20,000 ft.-lbs. of work? 

Prob. 7. On entering a heating oven cold air expands to twice its volume. Wha 
work is done per cubic foot of air? 

Prob. 8. A projectile is forced from a gun by a constant air pressure of 1000 Ibs. pe 
square inch. Before it begins to move there is J a cu.ft. of air in the barrel, and at the instant 
it leaves the barrel the volume is 10 cu.ft. What work was done on the projectile? 

Prob. 9. Water is forced from a tank against a head of 75 ft. by filling the tank witlj 
compressed air. How much work is done in emptying a tank containing 1000 cu.ft.? 

4. Work of Acceleration and Resultant Velocity. When a force acting 
on a mass is opposed by an equal resistance there may be no motion at all, oil 
there may be motion of constant velocity. Any differences, however, bet weeri 
the two opposing forces will cause a change of velocity so long as the difference] 
lasts, and this difference between the two forces may be itself considered as thcj 
only active force. Observations on unresisted falling bodies show that the$| 
increase in velocity 32.16 ft. per second for each second they are free to fall 
and this quantity is universally denoted by g. If then, a body have anj 
velocity, ui, and be acted on by a force equal to its own weight in the direction 
of its motion for a time, T seconds, it will have a velocity U2 after that time. 



It may be that the force acting is not equal to the weight of the body, in which 
case the acceleration will be different and so also the final velocity, due to the 
action of the force, but the force producing any acceleration will be to the 
weight of the body as the actual acceleration is to the gravitational acceleration, 
So that 

Actual acceleration force _ actual acceleration 

Weight of body or gravitational force gravitational acceleration (g) f 
and 

Actual accelerating force = - r - -. ~. - T-T X actual acceleration. 

gravitational acceleration (g) 

or 

change of velocity 



F= MX. . . . ........... (2) 

The work performed in accelerating a body is the product of the resistance 
met into the distance covered, L., while the resistance, or the above-defined force, 
acts, or while the velocity is being increased. This distance is the product of 
the time of action and the mean velocity, or the distance in feet, 

.......... (3) 



WORK AND POWER. 7 

The work is the product of Eqs. (2) and (3), or, work of acceleration is 

Jjr _M(u 2 -Ui) ^ 



where w is the weight in pounds. Exactly the same result will be obtained by 
the calculus when the acceleration is variable, so that Eq. (4) is of universal 
application. 

The work performed in accelerating a body depends on nothing but its mass 
and the initial and final velocities, and is in every case equal to the product of 
half the mass and the difference between the squares of the initial and final 
velocities, or the product of the weight divided by 6^-4 and the difference between 
the squares of the initial and final velocities. 

It frequently happens that the velocity due to the reception of work is desired, 
and this is the case with nozzle flow in injectors and turbines, where the steam 
performs work upon itself and so acquires a velocity. In all such cases the 
velocity due to the reception of the work energy is 



64.32TF 

U 2 = 



where W is work in foot-pounds and w, as before, is weight in pounds. Or if 
the initial velocity be zero, as it frequently is, 

U2 = l^= 1 64.32-. . . . \" . (6) 

\ w \ w 

For conversion of velocity units, Table VIII, at the end of the Chapter, 
is useful. 

Example. A force of 100 Ibs. acts for 5 seconds on a body weighing 10 Ibs.; if the 
original velocity of the body was 5 ft. per second, what will be the final velocity, the 
distance traveled and the work done? 



10 (M.-5). 
~32^~^5 ' 
U 2 = 1615 ft. per second; 



W= M(U1 * -"'> = 



405,000 ft.-lbs. 



Prob. 1. A stone weighing \ Ib. is dropped from a height of 1 mile. With what veloc- 
ity and in what length of time will it strike if the air resistance is zero? 

Prob. 2. A car moving 20 miles per hour and weighing 25 tons is brought to rest in 
500 ft. What is the negative acceleration, the time required to stop, and the work done? 



8 



ENGINEERING THERMODYNAMICS 



Prob. 3. Steam escapes through an opening with a velocity of half a mile per second. 
How many foot-pounds of energy were imparted to each pound of it to accomplish this? 

Prob. 4. A weight of 100 Ibs. is projected upward with a constant force of 200 Ibs. 
How much further will it have gone at the end of 10 seconds than if it had been merely 
falling under the influence of gravity for the same period of time? 

Prob. 6. A projectile weighing 100 Ibs. is dropped from an aeroplane at the height of 
J mile. How soon will it strike, neglecting air resis ance? 

Prob. 6. A water-wheel is kept in motion by a jet of water impinging on flat vanes. 
The velocity of the vanes is one-half that of the jet. The jet discharges 1000 Ibs. of 
water per minute with a velocity of 200 ft. per second. Assuming no losses, what is 
amount of the work done? 

Prob. 7. With the wind blowing 30 miles per hour, how much work could a 12-ft. 
windmill perform if 25 per cent of the available work were utilized. 

NOTE. The weight of a cubic foot of air may be taken as .075 Ib. 

Prob. 8. An engine has a piston speed of 600 ft. per minute and runs at 150 R.P.M. 
If the reciprocating parts weigh 500 Ibs., how much work is done in accelerating the 
piston during each stroke? 

Prob. 9. A flywheel with rim 10 ft. in diameter to center of section and weighing 5 
tons, revolves at a rate of 150 R.P.M. ; 100,000 ft.-lbs. of work are expended on it. How 
much will the speed change? 

5. Graphical Representation of Work. As work is always a product of 
force and distance or pressure and volume, it may be graphically expressed by 



S 3 



D 



A 12345 

Distances in Feet 
FIG. 1. Constant Force, Work Diagram, Force-Distance Coordinates. 

an area on a diagram having as coordinates the factors of the product. It is 
customary in such representations to use the horizontal distances for volumes 
and the vertical for pressures, which, if laid off to appropriate scale and 
in proper units, will give foot-pounds of work directly by the area enclosed. 
Thus in Fig. 1, if a force of 5 Ibs. (AB) act through a distance of 5 ft. (BC) 
there will be performed 25 foot-pounds of work as indicated by the area of the 



WOEK AND POWER 



B 




















p 
















































































































D 


A 1 2 3 i 5 



rectangle ABCD, which encloses 25 unit rectangles, each representing one foot- 
pound of work. 

If a steam cylinder piston suffers a displacement of 5 cu.ft. under the steam 
pressure (absolute) of 5 Ibs. per square foot then the operation which results 
in the performance of 25 foot-pounds of work is represented by the diagram 
Fig. 2, ABCD. 



a 

1 3 



Volumes iu Cubic Feet 
FIG. 2. Constant Pressure Work Diagram, Pressure- Volume Coordinates. 

Prob. 1. Following the method given for Fig. 1, draw a diagram for the example of 
Section 3 

Prob. 2. By means of a diagram, show that the work done by a pressure of 1000 Ibs. 
per square foot traversing a distance of 10 ft. is 10,000 ft.-lbs. 

Prob. 3. Draw a diagram for the case of a volume change from 1 to 10 cu.ft. while the 
pressure acting is 20 Ibs. per square inch. 

Prob. 4. Draw a pressure volume diagram for the case of forcing a piston out of a 
cylinder by a water pressure of 15,000 Ibs. per square foot, the volume of the cylinder at 
the start is \ cu.ft. and at the end 6 cu.ft. Make a diagram to scale and report work 
per square inch of diagram. 

Prob. 5. A pump draws in water at a constant suction pressure of 14 Ibs. and dis- 
charges it at a constant delivery pressure of 150 Ibs. per sq.in. Considering the pump 
barrel to be empty at beginning of suction and end of delivery and to contain 3 cu.ft. 
when full, draw the diagram for this case and find the foot-pounds of work done. 

Prob. 6. In raising a weight a man pulls on a rope with a constant force of 80 Ibs. 
If the weight is lifted 40 ft., find from a diagram the work done. 

Prob. 7. In working a windlass a force of 100 Ibs. is applied at the end of a 6-ft. 
lever, the drum of th windlass being 1 ft. in diameter. Draw a work diagram for 
work applied and for work done in lifting if there be no loss in the windlass. 

Prob. 8. The steam and water pistons of a pump are on the same rod and the area 
of the former is twice that of the latter, the stroke being 3 ft. Show by a diagram 
that the work done in the two cylinders is the same if losses be neglected. 

Prob. 9. An engine exerts a draw-bar pull of 8000 Ibs. at speed of 25 miles an hour. 
A change in grade occurs and speed increases to 40 miles per hour and the pull decreases 
to 5000 Ibs. Show by a diagram the change in horse-power. 



10 



ENGINEERING THERMODYNAMICS 



6. Work by Pressure Volume Change. Suppose that instead of being 
constant the pressure were irregular and, being measured at intervals of 1 cu.ft. 
displacement, found to be as follows: 



Pressure. 
Lbs. per Sq.Ft. 


Displacement 
Volume Cu.Ft. 


100 





125 


1 


150 


2 


100 


3 


75 


4 


50 


. 5 



150 












C ' 
















& 








.^ 


\ 
















lm 


A' 


E 


^^ 




\ 
















8 




^^ 








\ 














b 


^^ 




B' 






\ 


D 



























^x. 












o 














^^s^ 












e 
















\^ 


E 








.g 


















^\ 








8 50 




















^^V^ 


F 




& 




























H 




















3 




1 2 3 4 5 T 






Volumes in Cubic Feet 

FIG. 3. Work Diagram, Pressure-Volume Coordinates. 

Relations. 



Discontinuous Pressure- Volume 



This condition might be plotted as in Fig. 3, A, B, C, D, E, F, G, H. The 
work done .will be the area under the line joining the observation points. In 
the absence of exact data on the nature of the pressure variations between the< 
two observation points A and B, a variety of assumptions might be made as to 
the precise evaluation of this area, as follows: 

(a) The pressure may have remained constant at its original value for th( 
first cubic foot of displacement, as shown dotted A-B,' and then suddenly have 
risen to B. In this case the work done for this step would be 100 foot-pounds | 

(6) Immediately after the measurement at A the pressure may have riser! 
to A' and remained constant during displacement A' to B, in which case ttuj 
work done would be 125 foot-pounds. 

(c) The pressure may have risen regularly along the solid line AB, in whicl 

100+125, 



case the work area is a trapezoid and has the value 
pounds. 



XI = 112.5 foot 



WORK AND POWER 



11 



It thus appears that for the exact evaluation of work done by pressure 
volume change, continuous data are necessary on the value of pressure with 
respect to the volume. If such continuous data, obtained by measurement or 
otherwise, be plotted, there will result a continuous line technically termed the 
pressure-volume curve for the process. Such a curve for a pressure volume 
change starting at 1 cu.ft. and 45 Ibs. per square foot, and ending at 7 cu.ft., 
and 30 Ibs. per square foot, is represented by Fig. 4, A, B, C, D, E. 

The work done during this displacement under continuously varying pressure 
is likewise the area between the curve and the horizontal axis when pressures are 
laid off vertically, and will be in foot-pounds if the scale of pressures is pounds 
per square foot and volumes, cubic feet. Such an irregular area can be divided 
into small vertical rectangular strips, each so narrow that the pressure is sensibly 
constant, however much it may differ in different strips. The area of the 
rectangle is PAF, each having the width AF and the height P, and the work 



assures in Pounds per Square Foot 

5 g S g 8 




cz 

D 










^ 


B 


-^ 


*>s 


















/ 


X 










X 


*v 












A 


/ 














\ 


x 




























\ 


X 






























*** 





































E 

























SO 12 34 56 7 

Volumes in Cubic Feet 

FIG. 4. Work Diagram, Pressure-Volume Coordinates. Continuous Pressure- Volume 

; Relations, 

ea will be exactly evaluated if the strips are narrow enough to fulfill the 
nditions of sensibly constant pressure in any one. This condition is true only 
lor infinitely narrow strips having the width dV and height P, so that each has 
the area PdV and the whole area or work done is 



W= PdV. 



(7) 



This is the general algebraic expression for work done by any sort of continuous 
pressure volume change. It thus appears that whenever there are available 
sufficient data to plot a continuous curve representing a pressure volume change, 
the work can be found by evaluating the area lying under the curve and bounded 
by the curve coordinates and the axis of volumes. The work done may be 
found by actual measurement of the area or by algebraic solution of Eq. (7), 
which can be integrated only when there is a known algebraic relation between 
the pressure and the corresponding volume of the expansive fluid, gas or vapor. 

Prob. 1. Draw the diagrams for the following cases : (a) The pressure in a cylinder 
12 ins. in diameter was found to vary at different parts of an 18-in. stroke as follows: 



12 



ENGINEERING THERMODYNAMICS 



Pressure in Pounds 
per Sq.In. 


Per Cent of 
Stroke. 


100 





100 


10 


100 


30 


100 


50 


83.3 


60 


71.5 


70 


62.5 


80 


55.5 


90 


50.0 


100 



(6) On a gas engine diagram the following pressures were found for parts of stroke. 



IN 



OUT 



V 


P 


V 


P 


V 


P 


0.25cu.ft. 


14.7 


0.1 


45.2 


0.13 


146.2 


0.20 " 


19.5 


0.102 


79.7 


0.15 


116.7 


0.14 " 


29.7 


0.104 


123.2 


0.17 


95.7 


0.10 " 


45.2 


0.106 


157.7 


0.19 


80.7 






0.108 


181.7 


0.21 


68.7 






0.11 


188.2 


0.23 


58.7 






0.12 


166.2 







. 



Prob. 2. Steam at a pressure of 100 Ibs. per square inch absolute is admitted to a 
cylinder containing .1 cu.ft. of steam at the same pressure, until the cylinder contains 
1 cu.ft., when the supply valve closes and the volume increases so that the product of 
pressure and volume is constant until a pressure of 30 Ibs. is reached. The exhaust 
valve is opened, the pressure drops to 10 Ibs. and steam is forced out until the volume 
becomes 1 cu.ft., when the exhaust valve closes and the remaining steam decreases in 
volume so that product of pressure and volume is constant until the original point 
is reached. Draw the pressure volume diagram for this case. 

Prob. 3. During an air compressor stroke the pressures and volumes were as 
follows : 



Volume in 
Cu.Ft. 


Pressure in Lbs. 
Sq.In. 


2.0 


14.0 


1.8 


15.5 


1.6 


17.5 


1.4 


20.0 


1.2 


23.3 


1.0 


28.0 


0.8 


28.0 


0.4 


28.0 


0.0 


28.0 



Draw the diagram to a suitable scale to give work area in foot-pounds directly. 
Prob. 4. Draw the diagrams for last two problems of Section 3. 



WOEK AND POWEE 13 

7. Work of Expansion and Compression. Any given quantity of gas or 
vapor confined and not subject to extraordinary thermal changes such as 
explosion, will suffer regular pressure changes for each unit of volume change, 
or conversely, suffer a regular volume change for each unit of pressure change, 
so that pressure change is dependent on volume change and vice versa. When 
the volume of a mass of gas or vapor, Vi, is allowed to increase to 2 by the 
movement of a piston in a cylinder, the pressure will regularly increase or 
decrease from PI to Pz, and experience has shown that no matter what the gas 
or vapor or the thermal conditions, if steady, the volumes and pressures will 
have the relation for the same mass,. 



(8) 



or the product of the pressure and s power of the volume of a given mass 
will always be the same. The exponent s may have any value, but usually 
lies between 1 and 1.5 for conditions met in practice. 

I The precise value of s for any given case depends on 
(a) The substance. 
(6) The thermal conditions surrounding expansion or compression, s being 
different if the substance receives heat from, or loses heat to, .external sur- 
roundings, or neither receives nor loses. 

(c) The condition of vapors as to moisture or superheat when vapors are 
under treatment. 

Some commonly used values of s are given in Table X at the end of this 
chapter for various substances subjected to different thermal conditions dur- 
ing expansion or compression. 

Not only does Eq. (8) express the general law of expansion, but it likewise 
expresses the law of compression for decreasing volumes in the cylinder with 
corresponding rise in pressure. Expansion in a cylinder fitted with a piston 
is called balanced expansion because the pressure over the piston area is 
balanced by resistance to piston movement and the mass of gas or vapor is 
substantially at rest, the work of expansion being imparted to the piston and 
resisting mechanism attached to it. On the other hand when the gas or vapor 
under pressure passes through a nozzle orifice to a region of lower pressure the 
falling pressure is accompanied by increasing volumes as before, but the work 
of expansion is imparted not to a piston, because there is none, but to the fluid 
itself, accelerating it until a velocity has been acquired as a resultant of the 
work energy received. Such expansion is termed free expansion and the law of 
P]q. (8) applies as well to free as to balanced expansion. This equation, then, 
is of very great value, as it is a convenient basis for computations of the work 
done in expansion or compression in cylinders and nozzles of all sorts involv- 
ing every gas or vapor substance. Some expansion curves for different values 
of s are plotted to scale in Fig. 5, and the corresponding compression curves in 
Fig. 6, in which 

Curve A has the exponent s= 
Curved " " s= .5 



14 



ENGINEERING THERMODYNAMICS 



Curve C has the exponent s = 1.0 



Curve D 
Curved 
Curved 
Curved 
Curved 



s = l.l 
8=1.2 
8 = 1.3 
s = 1.4 
s = 1.5 



21000 



1 2 34 5 6 7 8 9 10 11 







FIG. 5. Comparison of Expansion Lines having Different Values'of s. 
The volume after expansion is given by 



so that the final volume depends on the original volume, on the ratio of the two 
pressures and on the value of the exponent. Similarly, the pressure after 
expansion 



WORK AND POWER 



15 



epends on the original pressure, on the ratio of the two volumes and on 
be exponent. 

The general equation for the work of expansion or compression can now be 
itegrated by means of the Eq. (8), which fixes the relation between pressures 
nd volumes. From Eq. (8), 




9 10 11 12 13 14 15 16 17 18 1'J 20 

Volumes in Cubic Feet 



FIG. 6. Comparison of Compression Curves having Different Values of s. 
which, substituted in Eq.(7), gives 



but as K is a constant, 



V s ' 



W=K 



(11) 



The integral of Eq. (11) will have two forms: 

(1) When s is equal to one, in which case 

(2) When s is not equal to one. 



P 2 V 2 = K l ; 



16 ENGINEERING THERMODYNAMICS 

Taking first the case when s is equal to one, 

2F 
F 



Whence 



W 



(a) 



(6) 



^ (c) 



^lo&g (d) 



When s = 






(12) 



Eqs. (12) are all equal and set down in different forms for convenience ir 

computation; in them 

T 7 2 = largest volume = initial vol. for compression = final vol. for expansion 
P2 = smallest pressure = initial pres. for compression = final pres. for expansion 
Fi = smallest volume = final vol. for compression = initial vol. for expansion 
PI = largest pressure = final pres. for compression = initial pres. for expansion 
These Eqs. (12) all indicate that the work of expansion and compression Q\ 

this class is dependent only on the ratio of pressures or volumes at the beginning 

and end of the process, and the PV product at either beginning or end, this 

product being of constant value. 

When the exponent s is not equal to one, the equation takes the form, 



rv t jy rv 2 
= K\ jT=K\ V~ s dV 
Jv V s Jv 



As s is greater than one, the denominator and exponents will be negative, so 
changing the form to secure positive values, 



JL/i L\ 

s-lVFi'- 1 TV" 1 /' 



This can be put in a still more convenient form. Multiplying and dividing aj 



1 1 

or 



7 2 ,-i FlS -i- 



v 2 



WORK AND POWER 
ubstituting the value of K = P 2 V 2 s = PiVi s , 

Whence 



17 



<> 






Whens^l, 



. (13) 



Eqs. (13) gives the work for this class of expansion and compression in terms 
pressure ratios and volume ratios and in them 

V 2 = largest volume = initial vol. for compression = final vol. for expansion; 
P 2 = smallest pressure = initial pres. f or compression = final pres. for expansion ; 
Vi = smallest volume = final vol. for compression = initial vol. for expansion; 
PI = largest pressure = final pres. for compression = initial pres. for expansion. 

The work of expansion or compression of this class is dependent according 
to Eqs. (13), upon the ratio of pressures or volumes at beginning and end of the 
process, the exponent, and on the pressure volume product appropriately taken. 
It should be remembered that for the result to be in foot-pounds appropriate 
units should be used and all pressures taken absolute. Examination of Eqs. 
(12) and (13), for the work done by expansion or compression of both classes, 
shows that it is dependent on the initial and final values of pressures and volumes 
and on the exponent s, which defines the law of variation of pressure with 
volume between the initial and final states. 

Example 1. Method of calculating Diagram, Figs. 5 and 6. Consider the curve 
for which s = 1.4 as typical of the group. 



Assumed Data. Vi = 1 .0 cu.ft. PI = 26,000 Ibs.. per square foot. 



Then 



PiVi s =K = 20,000 Xl 1 ' 4 =20,000. 



For any other value of P, V was found from the relation, 

r-rr' 



.715 



18 ENGINEERING THERMODYNAMICS 

Let P x = 6000, 
then V x = 



or 



.715 



log 3.33 = .5224 

.715 X. 5224 = .373= log V t . 
/. F*=2.36. 

A series of points, as shown below, were found, through which the curve was drawn 



p 


20,000 
P ' 


20,000 
log-; p - 


1 20,000 
T log p-. 


V 


18000 


1.111 


0.0453 


0.032 


1.08 


14000 


1.430 


0.1553 


0.111 


1.30 


10000 


2.000 


0.3010 


0.214 


1.64 


6000 


3.330 


0.5224 


0.373 


2.36 


2000 


10.000 


1.0000 


0.714 


5.18 


1000 


20.000 


1.3010 


0.930 


8.51 



Curves for other values of s were similarly drawn. Starting at a common volume 
of 20 cu. ft. the compression curves of Fig. 6 were determined by the same methods 

Example 2. A pound of air at 32 F. and under atmospheric pressure is compressed 
to a pressure of five times the original. What will be the final volume and the wort 
done if s = l and if s = 1.4? The volume of 1 Ib. of air at 32 F. and one atmosphere 
is 12.4 cu.ft. approx. 



Fors = l, 



Pi - 

PT 5 ' 



12.4cu.ft.; 



Z?=^ = ^4=5, whence 

Vl /2 Vl 



! =2.48 cu.ft. 



=P 2 F 2 log e = 2116xl2.41oge5; 

f 2 

= 21 16 X 12.4 X 1 .61 = 42,300 foot-pounds. 



For s = 1.4, 



7, _ /PA O 



il^-*^) 



5 may be raised to the .71 power by means of logarithms as follows: (5)' 71 is equal to 
the number whose logarithm is .71 log 5. 



WORK AND POWER 19 

Log 5 =.699, .71 X.699 =.4963, and number of which this is the logarithm is 3.13, 

icnce, 

Fi = F 2 4-3.13 or Fi=3.96; 



2116X12.4 ft _ lbg 



The value of W can also be found by any other form of equation (13) such as, 



The value of Vi being found as before, the work expression becomes after numerical 
substitution 



w 10,580X3.96 [ /3.96y 4 ] 
.4 ~\12.4/ 



As the quantity to be raised to the .4 power is less than one, students may find it 
easier to use the reciprocal as follows: 



/3.96V4 
\12.4/ 



\3.96/ 



08 ' ' 632 



y ,10,580X3.96 (1 _ 

Prob. 1. Find V l and W for Example 2 if s =1.2 and 1.3. 

Prob. 2. If a pound of air were compressed from a pressure of 1 Ib. per square inch 
absolute to 15 Ibs. per square inch absolute find Vi and W when s = 1 and 1.4. F 2 = 180 
cu.ft. What would be the H.P. to compress 1 Ib. of air per minute? 

Prob. 3. Air expands so that s = l. If Pi = 10,000 Ibs. per square foot, Fi = 10 cu.ft. 
jind F 2 = 100 cu.ft. and the expansion takes place in 20 seconds, whaj is the H.P. devel- 
oped? 

Prob. 4. 100 cu.ft. of air at atmospheric pressure is compressed in a cylinder to a pres- 
sure of 8 atmospheres and then expelled against this constant pressure. Find graphically 
and by calculation the foot-pounds of work done for the case where s = 1 and for the case 
where s = 1.4. 

Prob. 5. At an altitude of 4000 ft., a r is compressed to a pressure of 60 Ibs. per 
sq.in. gage. Find .he H.P. required to compress 1000 cu.ft. of free air per minute. 

Prob. 6. From the algebraic equation show how much work is done for a volume 
change of 1 to 4, provided pressure is originally 1000 Ibs. per square foot when 

(a) PW-&, 
(6) PV=K 2 , 
(c) PV*=K 3 . 



20 ENGINEERING THERMODYNAMICS 

Prob. 7. A vacuum pump compresses air from 1 Ib. per square inch absolute to 15 
Ibs. per square inch absolute and discharges it. An air compressor compresses air from 
atmosphere to 15 atmospheres and discharges it. Compare the work done for equal 
initial volumes, s = 1 .4. 

Prob. 8. For steam expanding according to the saturation law, compare the work 
done by 1 Ib. expanding from 150 Ibs. per square inch absolute to 15 Ibs. per square inch 
absolute with, the work of the same quantity expanding from 15 Ibs. to 1 Ib. per square 
inch absolute. 

NOTE. 1 Ib. of steam occupies 3 cu.ft. at 150 Ibs. per square inch absolute. 

Prob. 9. Two air compressors of the same size compress air adiabatically from atmos- 
phere to 100 Ibs. gage and discharge it. One is at sea level, the other at 10,000 ft. ele- 
vation. Compare the work in the two cases. 

8. Values of Exponent s Defining Special Cases of Expansion or Compres- 
sion. There are three general methods of finding s for the definition of particular 
cases of expansion or compression to allow of the solution of numerical problems. 
The first is experimental, the second and third thermodynamic. If by measure- 
ment the pressures and volumes of a series of points on an expansion or com- 
pression curve, obtained by test with appropriate instruments, for example, 
the indicator, be set down in a table and they be compared in pairs, values of 
s can be found as follows: Calling the points A, B, C, etc., then, 



and 

log Pa~\-S log Fa = log Pb + S log Vb, 

or 

(log 76 -log 7a) =log Pa-log Pb, 

hence 

_logP.-logP., 

~logF,-logF. 
or 

(14) 



log 



\ 
log ft 



According to Eq. (14a), if the difference between the logarithms of the pressures 
at B and A be divided by the differences between the logarithms of the volumes 
at A and B respectively, the quotient will be s. According to Eq. (146), the 
logarithm of the ratio of pressures, B to A , divided by the logarithm of the ratio 
of volumes, A to B respectively will also give s. It is interesting to note that 
if the logarithms of the pressures be plotted vertically and logarithms of volumes 
horizontally as in Fig. 7, then the line AC equal to the intercept on the horizontal 
axis represents the difference between the logarithms of volumes or, 



WORK AND POWER 



21 



; nd similarly 
lence 



- log P a - 






CB 

= = tana. 

CA 



>r the slope of the line indicates the value of s. This is a particularly valuable 

nethod, as it indicates at a glance the constancy or variability of s, and there 

ire many cases of practice where s does vary. Should s be constant the line 

Lvill be straight; should it be variable the line will be curved, but can generally 



\ 



LloffVA- 



LogVe 



BC 



.0 .2 .4 .6 .8 1. 

Log. V 
FIG. 7. Graphic Method of Finding s, from Logarithms of Pressures and Volumes. 

be divided into parts, each of which is substantially straight and each will 
have a different s. It is sometimes most convenient to take only the beginning 
and end of the curve and to use the value of s corresponding to these points, 
neglecting intermediate values. 

A second method for finding s for a given compression or expansion line 
by means of areas is indicated in a note in Section 17 of this Chapter that 
is omitted here because it depends on formulas not yet derived. It is by this 
sort of study of experimental data that most of the valuable values of s have 
been obtained. There is, however, another method of finding a value for 
s by purely thermodynamic analysis based on certain fundamental hypo- 
theses, and the value is as useful as the hypotheses are fair or true to the 
facts of a particular case. 

One of the most common hypotheses of this sort is that the gas or vapor 
undergoing expansion or compression shall neither receive any heat from, 
nor give up any to bodies external to itself during the process, and such a process 
is given the name adiabatic. Whether adiabatic processes are possible in actual 
cylinders or nozzles does not affect the analysis with which pure thermody- 
namics is concerned. By certain mathematical transformations, to be carried 
out later, and based on a fundamental thermodynamic proposition, the adia- 



22 ENGINEERING THERMODYNAMICS 

batic hypothesis will lead to a value of s, the use of which gives results valuable 
as a basis of reference, and which when compared with an actual case will per- 
mit of a determination of how far the real case has departed from the adiabatic 
condition, and how much heat has been received or lost at any part of the 
process. The particular value of s which exists in an adiabatic change is repre- 
sented by the symbol y. 

Another common hypothesis on which another value of s can be derived, 
is that gases in expansion or compression shall remain at a constant temperature 
thus giving rise to the name isothermal. This is generally confined to gases 
and superheated vapors, as it is difficult to conceive of a case of isothermal or con- 
stant temperature expansion or compression of wet vapors, as will be seen later.] 

In the study of vapors, which, it must be understood, may be dry or wet 
that is, containing liquid, a common hypothesis is that during the expansion 01 
compression they shall remain just barely dry or that they shall receive or lost 
just enough heat to keep any vapor from condensing, or but no more thai 
sufficient to keep any moisture that tends to form always evaporated. Expan- 
sion or compression according to this hypothesis is said to follow the saturatior 
law, and the substance to remain saturated. It will appear from this therma 
analysis later that the value of s for the isothermal hypothesis is the same for al 
gases and equal to one, but for the adiabatic hypothesis s=y will have 
different value for different substances, though several may have the sarm 
value, while for vapors y will be found to be a variable for any one, its valu< 
depending not only on the substance, but on the temperatures, pressures anc 
wetness. 

When gases or vapors are suffered to expand in cylinders and nozzles o: 
caused to compress, it is often difficult and sometimes impossible or perhapn 
undesirable to avoid interference with the adiabatic conditions for vaporr 
and gases, with the isothermal for gases or with the saturation law for vapors 
yet the work to be done and the horse-power developed cannot be predictecj 
without a known value of s, which for such cases must be found by expori 
ence. A frequent cause of interference with these predictions, which shoukj 
be noted, is leakage in cylinders, which, of course, causes the mass undo. 
treatment to vary. 

According to these methods those values of s have been found which arc 
given in Table X, at the end of the Chapter. Mixtures of common gases sucli 
as constitute natural, producer, blast furnace or illuminating gas, alone o.l 
with air or products of combustion, such as used in internal combustion enginos 
have values of s that can be calculated from the elementary gases or measurecj 
under actual conditions. 

All vapors, except those considerably overheated, have variable exponent; 
for adiabatic expansion and compression. This fact makes the exact solutioi 
of problems of work for wet vapors, expanding or compressing, which form th 
bulk of the practical cases, impossible by such methods as have been described 
This class of cases can be treated with precision only by strictly therma 
methods, to be described later. 



WORK AND POWER 



23 



Prob. 1. By plotting the values for the logarithms of the following pressures and vol- 
mes, see if the value for s is constant, and if not find the mean value in each case. 



V 

10 

11 

12 



11 
12 



(a) GAS ENGINE COMPRESSION 

p V p V 

45.2 13 32.2 18 

39.7 14 29.7 20 

35.7 16 24.7 25 

(6) GAS ENGINE EXPANSION 

p V p V 



188.2 
166.2 



13 
15 
17 



146.2 

116.7 

65.7 



19 
21 
23 



P 

21.0 
19.5 
14.7 



P 

80.7 

68.7 
58.7 



V 

2.242 
2.994 
4.556 



(c) STEAM EXPANSION 
p V p 



203.3 

145.8 

89.9 



7.338 52.5 
12.44 28.8 
22.68 14.7 



ob. 2. By plotting the values for the logarithms of the volumes and pressures on 
he expansion and compression curves of the following cards, find value for s. 



Atmosphere 




A.tmosi 



24 



ENGINEERING THERMODYNAMICS 



100 

140 




Atmosphere 




Atmosphere 



Prob. 3. From the steam tables at the end of Chapter IV. select the pressures ari 
volumes for dry-saturated steam and find the value of s between 

(a) 150 Ibs. per square inch and 1 Ib. per square inch. 
(6) 15 

Prob. 4. Find for superheated steam at 150 Ibs. per square inch and with 10( 
of superheat expanding to 100 Ibs. per square inch without losing any superheat, tl 
corresponding value of s, using tabular data. 

Prob. 5. From the ammonia table data for dry-saturated vapor find the value of 
between 

(a) 150 Ibs. per square inch and 1 Ib. per square inch. 
(6) 15 

9. Work Phases and Cycles, Positive and Negative and Net Work. ACCOM 
ing to the preceding it is easy to calculate or predict numerically the work 
expansion or compression whenever the conditions are sufficiently definite \ 
permit of the selection of the appropriate s. It very seldom happens, howev* 
that the most important processes are single processes or that the work f 
expansion or compression is of interest by itself. For example, before expansic| 
can begin in a steam cylinder steam must be first admitted, and in air coi 
pressors air must be drawn in before it can be compressed. Similarly, aft! 
expansion in a steam cylinder there must be an expulsion of used vapor befos 
another admission and expansion can take place, while in the air compress i 1 
after compression the compressed air must be expelled before more can ent|? 
for treatment. The whole series of operations is a matter of more concei 
than any one alone, and must be treated as a whole. The effect can be me 5 
easily found by the summation of the separate effects, and this method : 
summation will be found of universal application. 

The whole series of processes taking place and involving pressure volur; 
changes is called a cycle, any one of them a phase. It is apparent that thdi 
can be only a limited number of phases so definite as to permit of the matK 



WOEK AND POWER 25 

atical treatment necessary for prediction of work, but it is equally clear that 
there may be a far greater number of combinations of phases constituting 
cycles. Before proceeding to analyze the action of steam or gas in a cylinder 
it is necessary first to determine on structural, thermal or any other logical 
grounds, what series of separate processes will be involved, in what order, and 
the pressure volume characteristics of each. Then and then only, can the 
cycle as a whole be treated. These phases or separate and characteristic proc- 
esses affecting the work done or involving pressure volume changes are divisible 
into two classes so far as the causes producing them are concerned, the first 
: thermal and the second mechanical. It requires no particular knowledge of 
thermodynamics to realize that if air be confined in a cylinder with a free piston 
and is heated, that the volume will increase while pressure remains constant, 
since the piston will move out with the slightest excess of pressure inside over 
what is outside. This is a pressure constant, volume increasing, phase, and 
. is thermal since it is a heat effect. If an ample supply of steam be available 
; from a boiler held at a constant pressure by the manipulation of dampers and 
|: fires by the fireman and the steam be admitted to a cylinder with a piston, 
i! the piston will move out, the pressure remaining constant and volume increas- 
ing. This is also a pressure constant, volume increasing phase, exactly as before, 
but is mechanical because it is due to a transportation of steam from the boiler 
to the cylinder, although in another sense it may be considered as thermal if 
the boiler, pipe and cylinder be considered as one part during the admission. 
A similar constant pressure phase will result when a compressor piston is 
forcibly drawn out, slightly reducing the pressure and permitting the outside - 
atmosphere to push air in, to follow the piston, and again after compression of 
air to a slight excess, the opening of valves to storage tanks or pipe lines having 
a constant pressure will allow the air to flow out or be pushed out of the cylinder 
at constant pressure. These two constant pressure phases are strictly mechan- 
ical, as both represent transmission of the mass. If a cylinder contain water 
and heat be applied without permitting any piston movement, there will be 
a rise of pressure at constant volume, a similar constant volume pressure rise 
phase will result from the heating of a contained mass of gas or vapor under the 
same circumstances, both of these being strictly thermal. 

However much the causes of the various characteristic phases may differ, 
the work effects of similar ones is the same and at present only work effects 
are under consideration. For example, all constant volume phases do no 
work as work cannot be done without change of volume. 

The consideration of the strictly thermal phases is one of the principal 
problems of thermodynamics, for by this means the relation between the work 
done to the heat necessary to produce the phase changes is established, and a 
basis laid for determining the ratio of work to heat, or efficiency. For the 
present it is sufficient to note that the work effects of any phase will depend 
only on the pressure volume changes which characterize it. 

Consider a cycle Fig. 8, consisting of (AB) t admission of 2 cu.ft of steam at 
a constant pressure of 200,000 Ibs. per square foot, to a cylinder originally 



26 



ENGINEERING THERMODYNAMICS 



containing nothing, followed by (BC), expansion with s = l, to a pressure o 
20,000 Ibs. per square' foot; .(CF), constant volume change of pressure, am 
(FG), constant pressure exhaust at 10,000 Ibs. per square foot. These opera- 
tions are plotted to scale. Starting at zero volume, because the cylindei 



200000 



456 8 5> 10 11 12 13 It 10 16 17 18 1 




Volumes in Cubic Feet 

FIG. 8. Analysis of Work Diagram for Admission Expansion and Exhaust of Engine without 

Clearance. 

originally contains nothing, and at a pressure of 200,000 Ibs. per square foot, 
the line AB, ending at volume 2 cu.ft., represents admission and the cross- 
hatched area under AB represents the 400,000 ft.-lbs. of work done during 
admission. At B the admission ceases by closure of a valve and the 2 cu.ft. 
of steam at the original pressure expands with lowering pressure according 
to the law 



PaV a = 



= 200,000 X 2 = 400,000 ft.-lbs., 



So that when 



= 4 cu.ft., P = - = 100,000 Ibs. per sq.ft.; 



= 5 cu.ft.. p = 40Q > OQQ = 
o 



V= 10 cu.ft., p = 



80,000 Ibs. per sq.ft.; 






This continues until F=20 at point C, at which time P=- =20,000 

^0 

Ibs. per square foot, and the work done during expansion is the cross-hatched 
area JBCD under the expansion curve BC, the value of which can be found by 
measuring the diagram or by using the formula Eq. (12), 



WOEK AND POWER 27 

hich on substitution gives 

W bc = 400,000 log e 10 = 400,000X2.3; 
= 920,000 ft.-lbs. 

'his completes the stroke and the work for the stroke can be found by addition 
f the numerical values, 

W ab = 400,000 ft.-lbs. ; 

W, c = 920,000 ft.-lbs.; 
W,a,+Wbc = 1,320,000 ft.-lbs. 

It is often more convenient to find an algebraic expression for the whole, 
ich for this case will be, 



= 400,000(1 +log e 10) =400,000X3.3 = 1,320,000 ft.-lbs. 

the return of the piston it encounters a resistance due to a constant pressure 
>f 10,000 Ibs. per square inch, opposing its motion; it must, therefore, do work 
n the steam in expelling it. Before the return stroke begins, however, the 
>ressure drops by the opening of the exhaust valve from the terminal pressure 
f the expansion curve to the exhaust or back pressure along the constant volume 
ine, CF t of course, doing no work, after which the return stroke begins, the 
>ressure volume line being FG and the work of the stroke being represented by 
he cross-hatched area DFGH, 






W fg = PfVf = 10,000 X 20 = 200,000 ft.-lbs. 



his is negative work, as it is done in opposition to the movement of the piston. 
The cycle is completed by admission of steam at constant zero volume, raising 
:he pressure along GA. The net work is the difference between the positive 
md negative work, or algebraically 



= 1,320,000-200,000= l,120,000~ft.-lbs. 



28 



ENGINEERING THERMODYNAMICS 



Consider now a cycle of an air compressor, Fig. 9. Admission or suctior 
is represented by AB, compression by BC, delivery by CD and constant volum* 
drop in pressure after delivery by DA. The work of admission is representec 
by the area ABFE or algebraically by 

Wab = PbVb, 

the work of compression by the area FBCG, or algebraically since s = lA b> 



15000 




7 8 9 10 11 12 13 It 15 16 17 18 19 20 

Volumes in Cubic Feet 



FIG. 9. Analysis of Work Diagram for Admission Compression and Delivery of Compresso] 

without Clearance. 

the work of delivery by the area CDEG, or algebraically 



The positive work is that assisting the motion of the piston during suction 
the area ABFE or algebraically PbVb. The negative work, that in opposi 
tion to the motion, is the sum of the compression and delivery work, the arei 
FBCDE, or algebraically, 



The net work is the difference and is negative, as such a cycle is mainly resistant 
and to execute it the piston must be driven with expenditure of work on the 
gas. The valus of the net work is, 



WOEK AND POWER 



29 



n expression which will be simplified in the chapter on compressors. This 
et work is represented by the area A BCD, which is the area enclosed by the 
ycle itself independent of the axes of coordinates. 

It might seem from the two examples given as if net work could be 
btained without the tedious problem of summation, and this is in a sense true 
the cycle is plotted to scale or an algebraic expression be available, but 
hese processes are practically equivalent to summation of phase results. It 
light also seem that the work area would always be that enclosed by the 
ycle, and this is true with a very important limitation, which enters when 
he cycle has loops. If, for example, as in Fig. 10, steam admitted A to B, 
xpanded along EC to a pressure C, then on opening the exhaust the pressure 
astead of falling to the back pressure or exhaust line as in Fig. 8, would here 




8 9 10 11 12 13 11 15 16 17 18 19 20 

Volumes in Cubic Feet 



10. Analysis of Work Diagram for Engine with Over-expansion Negative Work Loop. 

ise along CD, as the back pressure is higher than the terminal expansion pres- 
ure, after which exhaust will take place at constant back pressure along DE. 
The forward stroke work is that under AB and BC or ABC EG, the return 
troke work is the area DEGH and the net work is 






Area ABC EG -Area DEGH. 



is the area HGECX is common to both terms of the difference, the net work 
y be set down as equal to 

Are&ABXH-CDX. 



It may be set down then in general for looped cycles that the net work area 
8 the difference between that of the two loops. If, however, the method laid 
lown for the treatment of any cycle be adhered to there need not be any dis- 
inction drawn between ordinary and looped cycles, that is, in finding the work 



30 



ENGINEERING THERMODYNAMICS 



of a cycle divide it into characteristic phases and group them into positive an 
negative, find the work for each and take the algebraic sum. 

Special cases of cycles and their characteristics for steam compressors an 
gas engine cylinders, as well as nozzle expansion, will be taken up later in mor 
detail and will constitute the subject matter of the next two chapters. 



Example 1. Method of calculating Diagram, Fig. 8. 

Va =0 CU.ft. 

Assumed data 



To obtain point C. 



v d =v c " 

V e =2 " 
5=1 



P C F C =P 6 F 6 or 



P a =200,000 Ibs. per square foot. 

P 6 =Pa 

PC =20,000 
P f = 10,000 

Pe=Pf. 



P & F 6 200,000X2 9n 

c =-p7" 20,000 =20 ' 



/. F c =20 and P c = 20,000. 

Intermediate points B to C are obtained by assuming various pressures an 
finding the corresponding volumes as for F c . 

Example 2. Method of calculating Diagram, Fig. 9. 



Assumed data 



F a =0cu.ft. 
F 6 =20 " 
V d =0 " 



To obtain point C, 



P a =21 16 Ibs. per square foot. 
P & =P 
P c = 14,812 " 
( P d =P c 



or = 



.715 



~=7, Iog7=.845 ; and .715 X. 845= log (^Y 4 =. 6105, 



or 
Therefore, 



F c =4.02. 
F c =4.02, and P c = 14,812. 



Intermediate values BtoC may be found by assuming pressures and finding volumes C( 
responding as for V c . 

Prob. 1. Steam at 150 Ibs. per square inch absolute pressure is admitted into a cyli 
der in which the volume is originally zero until the volume is 2 cu.ft., when the valve 
closed and expansion begins and continues until the volume is 8 cu.ft., then exhai 
valve opens and the pressure falls to 10 Ibs. absolute and steam is entirely swept 01 
Draw the diagram and find the net work done. 



WORK AND POWEE 31 

Prob. 2. A piston moving forward in, a cylinder draws in 10 cu.ft. of C0 2 at a pressure 
f .9 of an atmosphere at sea level and then compresses it adiabatically until the pressure 
ises to 9 atmospheres and discharges it at constant pressure. Draw the diagram and 
jid net work done. 

Prob. 3. A cylinder 18 ins. in diameter and 24 ins. piston stroke receives steam at 

00 Ibs. per square inch absolute pressure for | of the stroke. It then expands to the 
nd of the stroke and is exhausted at atmospheric pressure. Draw the diagram and 
nd the H.P. if the engine makes 100 strokes per minute. 

Prob. 4. Two compressors without clearance each with a cylinder displacement of 

cu.ft. draw in air at 14 Ibs. per square inch absolute and compress it to 80 Ibs. per 

quare inch absolute before delivery. Find the difference in H.P. per 1000 cu.ft. of 

*ee air per minute if one is compressing isothermally and the other adiabatically. 

)raw diagram for each case. 

Prob. 5. A quantity of air 5 cu.ft. in volume and at atmospheric pressure is compressed 

1 a cylinder by the movement of a piston until the pressure is 50 Ibs. per square inch 
age. If the air be heated the pressure will rise, as in an explosion. In this case the 
iston remains stationary, while the air is heated until the pressure reaches 200 Ibs. per 
}uare inch gage. It then expands adiabatically to the original volume when the 
ressure is reduced to atmosphere with no change in volume. Draw the diagram, and 
nd the work done. 

Prob. 6. The Brayton cycle is one in which gas is compressed adiabatically and then, 
y the addition of heat, the gas is made to expand without change of pressure. Adi- 
batic expansion then follows to original pressure and the cycle ends by decrease in volume 
) original amount without change of pressure. Draw such a cycle starting with 5 cu.ft. 
: air at atmospheric pressure, compressing to 4 atmospheres, expanding at constant 
ressure to 5 cu.ft., expanding adiabatically to original pressure and finally ending at 
iginal point. Find also, work done. 

Prob. 7. In the Ericsson cycle air is expanded at constant temperature, cooled at 
mstant pressure, compressed at constant temperature and receives heat at constant 
)lume. Draw a diagram for the case where 5 cu.ft. at atmospheric pressure are com- 
-essed to 1 cu.ft., heated until volume is 8 cu.ft., expanded to atmosphere and then 
>oled to original volume. Find the work. 

Prob. 8. In the Stirling cycle constant volume heating and cooling replace that at 
tnstant pressure in the Ericsson. Draw diagram starting with 5 cu.ft. and atmsopheric 
essure compressing to 1 cu.ft. and then after allowing the pressure to double, expand 
i original volume and cool to atmosphere. Find the work. 

Prob. 9. The Joule cycle consists of adiabatic compression and expansion and con- 
ant pressure heating and cooling. Assuming data as in last problem draw the 
agram and find the work. 

Prob. 10. The Carnot cycle consists of isothermal expansion, adiabatic expansion, 
)thermal compression and adiabatic compression. Draw the diagram for this cycle 
id find the work. 

10. Work Determination by Mean Effective Pressure. While the methods 
ready described are useful for finding the .work done in foot-pounds for a defined 
cle with known pressure and volume limits, they are not, as a rule, convenient 
r the calculation of the work done in a cylinder of given dimensions. As 
prk done can always be represented by an area, this area divided by its length 
U give its mean height. If the 'area be in foot-pounds with coordinates 



32 ENGINEERING THERMODYNAMICS 

pounds per square foot, and cubic feet, then the division of area in foot-poundi 
by length in cubic feet will give the mean height or the mean pressure ii 
pounds per square foot. Again, dividing the work of the cycle into forward 
stroke work and back-stroke work, or the respective foot-pound areas divide< 
by the length of the diagram in cubic feet, will give the mean forward pressur 
and the mean back pressure. The difference between mean forward pressur 
and mean back pressure will give the mean effective pressure, or that averagi 
pressure which if maintained for one stroke would do the same work as thj 
cycle no matter how many strokes the cycle itself may have required for it ! 
execution, which is very convenient considering the fact that most gas engine) 
require four strokes to complete one cycle. The mean effective pressure mai 
also be found directly from the enclosed cycle area, taking proper account e 
loops, as representative of net work by dividing this net work area by the lengt 
of the diagram in appropriate units. This method is especially convenier 
when the diagram is drawn to odd scales so that areas do not give foot-pounc 
directly, for no matter what the scale the mean height of the diagram, whe 
multiplied by the pressure scale factor, represents the mean effective pressur; 
This mean height can always be found in inches for any scale of diagram hi 
finding the area of the diagram in square inches and by dividing by the lengti 
in inches, and this mean height in inches multiplied by the scale of pressur. 
in whatever units may be used will give the mean effective pressure in the san: 
units. 

Mean pressures, forward, back or effective, are found and used in two genen j 
ways; first, algebraically, and second graphically and generally in this ca? 
from test records. By the first method, formulas, based on some assumli 
laws for the phases, can be found, and the mean effective pressure and its val j 
predicted. This permits of the prediction of work that may be done by a giv i| 
quantity of gas or vapor, or the work per cycle in a cylinder, or finally the hon- 
power of a machine, of which the cylinder is a part, operating at a given spei 
and all without any diagram measurement whatever. By the second methcJ 
a diagram of pressures in the cylinder at each point of the stroke can be obtain 1 
by the indicator, yielding information on the scale of pressures. The net wcc| 
area measured in square inches, when divided by the length in inches, ghssl 
the mean height in inches, which, multiplied by the pressure scale per injij 
of height, gives the mean effective pressure in the same units, which 
usually pounds per square inch in practice. 

As an example of the algebraic method of prediction, consider the cyj 
represented by Fig. 8. The forward work is represented by 



Forward work = P b V b (l+\og e ~j , 

the length of the diagram representing the volume swept through in the 
formance of this work is V c , hence 

Mean forward pressure = .-( l + log c ~J. 



WORK AND POWER 33 

lut PbVb = P c V c by the law of this particular expansion curve, hence 
Mean forward pressure = P c (l+log e ^- c j. 

js the back pressure is constant its mean value is this constant value, hence 

Constant (mean) back pressure = P/. 
y subtraction 

Mean effective pressure = P C ( l+log e ^J P f 

= 3.3P c -P /; 

= 3.3X20,000-10,000; 

= 66,000-10,000 = 56,000 Ibs. per sq.ft. 

The work done in foot-pounds is the mean effective pressure in pounds per 
re foot, multiplied by the displacement in cubic feet. 

Tf = 56,OOOX20 = 1,120,000 ft.-lbs. as before. 




i(i. 11. Gas-Engine Indicator Card. For Determination of Mean Effective Pressure 

without Volume Scale. 

! As an example of the determination of mean effective pressure from a test 
1 indicator diagram of unknown scale except for pressures, and without axes 
p coordinates, consider Fig. 11, which represents a gas engine cycle in four 
jrokes, the precise significance of the lines being immaterial now. The 
pressure scale is 180 Ibs. per square inch, per inch of height. 
By measurement of the areas in square inches it is found that 

Large loop area CDEXC =2.6 sq.in. 
Small loop area ABXA =0.5 sq.in. 

pc Net cycle area =2.1 sq.in. 

Length of diagram . =3.5 in. 

Mean height of net work cycle = 0.6 in. 
Mean effective pressure = 1 20 X. 06 = 72 Ibs. per square inch. 



34 ENGINEERING THERMODYNAMICS 

It is quite immaterial whether this diagram were obtained from a large < 
a small cylinder; no matter what the size, the same diagram might be secure 
and truly represent the pressure volume changes therein. If this particul 
cylinder happened to have a diameter of 10 ins. and a stroke of 12 ins. tl 
work per stroke can be found. The area of the cylinder will be 78.54 sq.in; 
hence the average force on the piston is 72 Ibs. per square inch X78.54 sq.ins. 
5654.88 Ibs., and the stroke being 1 ft. the work per stroke is 5654.88 ft.-ftj 
Both of these methods are used in practical work and that one is adopted \ 
any particular case which will yield results by the least labor. 

Prob. 1. An indicator card from an air compressor is found to have an area of 3.1 
sq. ins., while the length is 2^ ins. and scale of spring is given as 80 Ibs. per square in! 
per inch height. What is m.e.p. and what would be the horse-power if the compress 
ran with a piston speed of 250 ft. per minute and had a piston 9 ins. in diameter? 

Prob. 2. For the same machine another card was taken with a 60-lb. spring and hj 
an area of 4.12 sq.ins. How does this compare with first card, the two having the sa:|f 
length? 

Prob. 3. A steam engine having a cylinder 18 ins. in diameter and a stroke of 24 ir 
takes in J cu.ft. of steam at 100 Ibs. absolute, allows it to expand and exhausts at atmt 
pheric pressure. An indicator card taken from the same engine showed a length oJ 
ins., an area of .91 sq.in. when an 80-lb. spring is used. How does the actual m.o 
compare with the computed? 

Prob. 4. Find m.e.p. by the algebraic method of prediction for, 

(a) Bray ton cycle; 

(6) Carnot cycle; 

(c) Stirling cycle; 

(d) Ericsson cycle; 

(e) Joule cycle. 

(See problems following Section 9). 

11. Relation of Pressure-Volume Diagrams to Indicator Cards. 7< 
Indicator. When a work cycle or diagram of pressure volume changes is drai 
to scale with pressures and volumes as coordinates, it is termed a press n 
volume or PV diagram, and may be obtained by plotting point by point ffm 
the algebraic expression for the law of each phase or by modifying the indicai 
card. The indicator card is that diagram of pressures and stroke obtai :K 
by applying the indicator to a cylinder in operation. This instrument cons fa 
essentially of a small cylinder in which a finely finished piston moves frm 
without appreciable friction, with a spring to oppose its motion, a pencil mechia- 
ism to record the extent of the motion, and a drum carrying paper which is mo fc 
in proportion to the engine piston movement. The indicator cylinder is ofci 
at the bottom and fitted with a ground union joint for attachment to the njto 
cylinder through a special cock, which when open permits all the varjta 
pressures in the main cylinder to act on the indicator piston, and when clcW 
to the main cylinder opens the indicator cylinder to the atmosphere. r f 



WORK AND POWER 35 

upper side of the indicator piston being always open to the atmosphere, its 

movement will be the result of the difference between the pressure in the main 

cylinder and atmospheric pressure. A helical spring, carefully calibrated and, 

therefore, of known scale, is fixed between the indicator piston and open cap or 

, head of its cylinder, so that whenever the pressure in the main cylinder exceeds 

atmosphere the indicator piston moves toward the open head of the indicator 

cylinder, compressing the spring. Pressures in the main cylinder if less than 

atmosphere will cause the indicator piston to move the other way, extending 

the spring. This compression and extension of the spring is found in the 

calibration of the spring to correspond to a definite number of pounds per 

square inch above or below atmosphere per inch of spring distortion, so that 

: the extent of the piston movement measures the pressure above or below atmos- 

,phere. A piston rod projects outward through the cylinder cap and moves a 

series of levers and links carrying a pencil point, the object of the linkage being 

to multiply the piston movement, but in direct proportion, giving a large 
P movement to the pencil for a small piston movement. A cylinder drum carry- 
ing a sheet of paper is pivoted to the cylinder frame so that the pencil move- 

} ment will draw on the paper a straight line parallel to the axis of the drum, if 
drum is stationary, or perpendicular to it if drum rotates and pencil is sta- 
tionary. The height of such lines then above or below a zero or datum line, 
which is the atmospheric line drawn with the cock closed, measures the pressure 
of the fluid under study. The springs have scale numbers which give the 
pressure, in pounds per square inch per inch of pencil movement. This paper- 
*'carrying drum is not fixed, but arranged to rotate about its axis, being pulled 
out by a cord attached to the piston or some connecting part through a pro- 
portional reducing motion so as to draw out the cord an amount slightly less 
than the circumference of the drum no matter what the piston movement. 
After having been thus drawn out a coiled spring inside the drum draws it back 
, on the return stroke. By this mechanism it is clear that, due to the combined 
movement of the pencil up and down, in proportion to the pressure, and that 
S| of the drum and paper across the pencil in proportion to the piston movement, 
a diagram will be drawn whose ordinates represent pressures above and below 
\ atmosphere and abscissae, piston stroke completed at the same time, or dis- 
placement volume swept through. It must be clearly understood that such 
j indicator diagrams or cards do not give the true or absolute pressures nor the 
rjtrue volumes of steam or gas in the cylinder, but only the pressures above or 
below atmosphere and the changes of volume of the fluid corresponding to the 
J piston movement. Of course, if there is no gas or steam in the cylinder at the 
rl beginning of the stroke, the true volume of the fluid will be always equal to the 
t . displacement, but no such cylinder can be made. 

While the indicator card is sufficient for the determination of mean effective 
1,5 pressure and work per stroke, its lack of axes of coordinates of pressure and 
^volume prevents any study of the laws of its curves. That such study is 

- important must be clear, for without it no data or constants such as the exponent 
1 8 can be obtained for prediction of results in other similar cases, nor can the 



36 



ENGINEERING THERMODYNAMICS 



presence of leaks be detected, or the gain or loss of heat during the various 
processes studied. In short, the most valuable analysis of the operations is 
impossible. 

To convert the indicator card, which is only a diagram of stroke or displace- 
ment on which are shown pressures above and below atmosphere into a pres- 
sure volume diagram, there must first be found (a) the relation of true or abso- 
lute pressures to gage pressures, which involves the pressure equivalent of 
the barometer, and (6) the relation of displacement volumes to true volume? 
of vapor or gas present, which involves the clearance or inactive volume of the 
cylinder. The conversion of gage to absolute pressures by the barometer 
reading has already been explained, Section 3, while the conversion of displace- 
ment volumes to true fluid volumes is made by adding to the displacement 
volume the constant value in the same units of the clearance, which is usually 
the result of irregularity of form at the cylinder ends dictated by structural 
necessities of valves, and of linear clearance or free distance between the pis- 
ton at the end of its stroke and the heads of the cylinder to avoid any possi-i 
bility of touching due to wear or looseness of the bearings. 



150 21600 




FIG. 12. Ammonia Compressor Indicator Card with Coordinates of Pressures and Volumes 
Added to Convert it into a Pressure- Volume Diagram. 

Let ABCD, Fig. 12, represent an indicator card from an ammonia compressor 
on which EF is the atmospheric line. The cylinder bore is 14 ins., stroke 
22 ins., and the scale of the indicator spring 100, barometer 28 ins., and measured 
clearance 32 cu.in. According to Table IX, 28 ins. of mercury corresponds tc 
13.753 Ibs. per square inch, and as 100 Ibs. per square inch, according to the spring 
scale, corresponds to 1 in. of height on the diagram, 1 Ib. per square inch cor- 
responds to 0.01 in. of height, or 13.75 Ibs. per square inch atmospheric pres- 
sure to .137 in. of height. The zero of pressures then on the diagram must! 
lie .137 in. below the line EF. Lay off then a line MH, this distance belo^l 
EF. This will be the position of the axis of volume coordinates. 

Actual measurement of the space in the cylinder with the piston at the enc I 
of its stroke gave the clearance volume of 32 cu. ins. As the bore is 141 
ins. the piston area is 153.94 sq. ins. which in connection with the stroke 



WORK AND POWER 37 

of 22 ins. gives a displacement volume of 22X153.94 = 3386.68 cu. ins. 

32 
Compared with this the clearance volume is .94 per cent of the 

displacement. It should be noted here that clearance is generally expressed 
in per cent of displacement volume. Just touching the diagram at the ends 
drop two lines at right angles to the atmospheric line intersecting the axis of 
volumes previously found at G and H . The intercept GH then represents the 
displacement, or 3386.68 cu.in. or 1.96 cu.ft. Lay off to the left of G, .0094, or 
in round numbers 1/100 of GH, fixing the point M, MG representing the clearance 
to scale, and a vertical through M the axis of pressures. The axes of coordinates 
are now placed to scale with the diagram but no scale marked thereon. The 
pressure scale can be laid off by starting at M and marking off inch points 
each representing 100 Ibs. per square inch. Pounds per square foot can also 
be marked by a separate scale 144 times as large. As the length of the diagram 
is 2.94 ins. and displacement 1.96 cu.ft., 1 in. of horizontal distance corresponds 
to .667 cu.ft. or 1 cu.ft to 1.50 ins. of distance. Lay off then from M dis- 
tances of 1.50 ins. and mark the first 1 cu.ft. and the second 2 cu.ft., dividing 
the intervals into fractions. A similar scale of volumes in cubic inches might 
also be obtained. 

Ity this process any indicator card may be converted into a pressure volume 
diagram for study and analysis, but there will always be required the two factors 
of true atmospheric pressure to find one axis of coordinates and the clearance 
volume to find the other. 

Prob. 1. If in cards Nos. 1 and 2, Section 8, the clearances are 5 per cent and 3 per 
cent respectively of the displacement, convert the cards to PV diagrams on the same 
base to scales of 4 ins. to 1 cu.ft. and 1 in. to 1000 Ibs. per square foot, for cylinders 
9j ins. and 14| ins. respectively in diameter and stroke 12 ins. 

Prob. 2. Do the same for cards Nos. 3 and 4, if clearances are 7 per cent and 4 per 
cent respectively, for cylinders 10 ins. and 17 ins. respectively in diameter and stroke 
12 ins. 

Prob. 3. Do the same for cards Nos. 7 and 8 if clearances are 12 per cent and 8 per 
3ent respectively and cylinders 12 ins. and 19 ins. in diameter and stroke 24 ins. 

12. To Find the Clearance. There are two general methods for the find- 
ing of clearance, the first a direct volumetric measurement of the space itself 
by filling with measured liquid and the second a determination by algebraic 
Dr graphic means from the location of two points on the expansion or com- 
pression curves of the indicator card based on an assumed law for the curves. 

The first method of direct measurement is the only one that offers even a 

oromise of accuracy, but even this is difficult to carry out because of the 

tendency of the measuring liquid to leak past piston or valves, which makes 

the result too large if the liquid be measured before the filling of the clearance 

space and too small if the liquid be measured after filling and drawing off. 

There is also a tendency in the latter case for some of the liquid to remain 

i inside the space, besides the possibility in all cases of the failure to completely 

; fill the space due to air pockets at high places. 



.100 



A" A' 



A A 



B"lB" 



B"|B\B 



40 



100 



10 



Displacement 
FIG. 13. 




40 



6 8 

Displacement 



10 



FIG. 14. 



A" 



A'\A 



20 



B' 



10 



1 

Displacement 
FIG. 15. 



WORK AND POWER 



39 



By the second general method any two points, A and B, on an expansion or 
mpression curve, Figs. 13, 14, 15, may be selected and horizontals drawn to 
e vertical line indicating the beginning of the stroke. The points A' and B' 
e distant from the unlocated axis an amount A'A" = B'B", representing the 
3arance. 

Let the clearance volume =CL; 
" the displacement up to A =D a ', 

the displacement up to B = D b ] 

the whole displacement = D ; 
" s be the exponent in PV S = constant, which defines the law of the curve. 

Then in general, 



= D a +Cl, 
= D b +Cl, 



/ i i\ i i 

\P a s-P b s ) = P b * D b - P a s D a 



Cl 
hence the clearance in whatever units the displacement may be measured will be 



Cl 



P b s D b -P a s 



] - ! * 



Cl 



id Cl, in per cent of the whole displacement will be 



D b /Pa\sD 



b_(' 
D \P h ) D 



Clearance as a fraction of displacement = c = 

,Pb 



len s = 1 this takes the form 



Clearance in fraction of displacement = c 



D 



. (15) 



40 ENGINEERING THERMODYNAMICS 

To use such an expression it is only necessary to measure off the atmospheij 
pressure below the atmospheric line, draw verticals at ends of the diagraj 
and use the length of the horizontals and verticals to the points in the formu j 
each horizontal representing one D and each vertical a P. 

Graphic methods for the location of the axis of pressures, and hence t'l 
clearance, depend on the properties of the curves as derived from analytid 
For example, when s=l, 



which is the equation of the equilateral hyperbola, a fact that gives a com 
name to the law, i.e., hyperbolic expansion or hyperbolic compression, 
common characteristics of this curve may be used either separately or toget 
the proof of which need not be given here, first that the diagonal of the rectan^e 
having two opposite corners on the curve when drawn through the other fr? 
corners will pass through the origin of coordinates, and second, that the otb 
diagonal drawn through two points of the curve and extended to intersect 11 
axes of coordinates will have equal intercepts between each point and t| 
nearest axis cut. 

According to the first principle, lay off, Fig. 16, the vacuum line or asfe 
of volume XY and selecting any two points A and B, construct the rectanjp 
ACBD. Draw the diagonal CDE and erect at E the axis of pressures E 1 9 
then will EZ and EY be the axes of coordinates. According to the seco.l 
principle, proceed as before to locate the axes of volumes XY and select t| 
points, A and B, Fig. 17. Draw a straight line through these points, whil 
represents the other diagonal of the rectangle ACBD, producing it to intil- 
sect XY at M and lay off AN = ~BM. Then will the vertical NE be the a; 
of pressures. It should be noted that these two graphic methods apply or 
when s = 1 ; other methods must be used when s is not equal to 1 . 

A method of finding the axis of zero volume is based upon the slope of t 
exponential curve, 

PF 5 =c. 

Differentiation with respect to V gives 



or 

Ps 



whence 

Ps 



-K-S) 



WORK AND POWER 



41 



In other words, the true volume at any given point on the known curve 
may be found by dividing the product of P and s by the tangent or the slope 
of the line at the given point, with the sign changed. This method gives 
results dependent for their accuracy upon the determination of the tangent to 
the curve, which is sometimes difficult. 



70 





Atraos 



heric 



Line 



357 
Displacement 

FIG. 16. 



70 



50 



30 





Atmosi 



heric Line 



5 7 

Displacement 

FIG. 17. 



11 



Graphic Methods of Locating the Clearance Line for Logarithmic or Hyperbolic Expansion 

and Compression Curves. 

The following graphical solution is dependent upon the principle just given, 
and while not mathematically exact, gives results so near correct that the 
-or is not easily measured. The curve ACB, Fig* 18, is first known experi- 
lentally or otherwise and therefore the value of s, and the axis FV from 
rhich pressures are measured is located. Assume that the axis of zero volume, 



42 



ENGINEERING THERMODYNAMICS 



KP, is not known but must be found. Selecting any two convenient points, 
A and B, on the curve, complete the rectangle AHBG with sides parallel and 
perpendicular to FV. The diagonal HG cuts the curve at C and the horizontal 
axis at E. From C drop the perpendicular CD. If now the distance DE 
be multiplied by the exponent s, and laid off DK, and the vertical KP erected, 
this may be taken as the zero volume axis. 

It cannot be too strongly stated that methods for the finding of clearance 
or the location of the axes of pressures from the indicator card, much as they 
have been used in practice, are inaccurate and practically useless unless it is 
positively known beforehand just what value s has, since the assumed value 



PI 




dV 



'-!>! 
ID 



FIG. 18. Graphic Method of Locating the Clearance Line for Exponential Expansion and 

Compression Curves. 



of s enters into the work, and s for the actual diagram, as already explained. 
is affected by the substance, leakage, by moisture or wetness of vapor and by. 
all heat interchange or exchange between the gas of vapor and its container. 



Prob. 1. If in card No. 6, Section 8, compression follows the law PV S =K 
where s = 1.4 and the barometer reads 29.9 ins. of mercury, locate the axes algebraically 
and graphically. 

Prob. 2. If in card No. 3, Section 8, expansion follows the law PV S =K 
where s = 1.37 and the barometer reads 29.8 ins. of mercury, locate the axes algebraically 
and graphically. 

Prob. 3. If in card No. 5, Section 8, expansion follows the law PV S =K 
where s = l and the barometer reads 27.5 ins. of mercury, locate axes algebraically ami 
graphically. 



WORK AND POWER 



43 



13. Measurement of Areas of PV Diagrams and Indicator Cards. Areas 

pressure volume diagrams or indicator cards must be evaluated for the 

termination of work or mean effective pressure, except when calculation by 

rmula and hypothesis is possible. There are two general methods applicable 

both the indicator card and PV diagram, that of average heights, and the 

animeter measure, besides a third approximate but very useful method, 

pecially applicable to plotted curves on cross-section paper. 

The third method assumes that the diagram may be divided into strips of 
: [ual width as in Fig. 19, which is very easily done if the diagram is plotted on 
oss-section paper. At the end of each strip, a line is drawn perpendicular to 
k le axis of the strip, such that the area intercepted inside the figure is apparently 
jual to that outside the figure. If this line is correctly located, the area of the 
:ct angular strip will equal the area of the strip bounded by the irregular lines. 



() 

50 
40 

: 

20 
10 








^**~ 





=^ 


\ 














^^ 




> 


7 








\ 


\ 














/ 

/ 












\ 




























\ 












I 














\ 





==*= 







^^- 






1 i 




1 








I I 



Volumes 

'IG. 19. Approximate Method of Evaluating Areas and Mean Effective Pressures of 
Indicator Cards and P.V. Diagrams. 



f the entire figure has irregular ends it may be necessary to subdivide one or 
>oth ends into strips in the other direction, as is done at the left-hand side of 
r ig. 19. The area of the entire figure will be equal to the summation of lengths 
>f all such strips, multiplied by the common width. This total length may be 
btained by marking on the edge of a strip of paper the successive lengths in 
uch a way that the total length of the strip of paper when measured will be the 
otal length of the strips. 

The mean height will be the total length of such strips divided by the num- 
)er of strip-widths in the length of the diagram. By a little practice the proper 
ocation of the ends of the strips can be made with reasonable accuracy, and 
consequently the results of this method will be very nearly correct if care is 
Jxercised. 

By dividing the diagram into equal parts, usually ten, and finding the length 
)f the middle of each strip, an approximation to the mean height of each strip 



44 



ENGINEERING THERMODYNAMICS 



will be obtained; these added together and divided by the number will grJ 
the mean height in inches from which the mean effective pressure may be founj 
by multiplying by the scale as above, or the area in square inches by multiphj 
ing by the length in inches, which can be converted into work by multiplyiD 
by the foot-pounds per square inch of area as fixed by the scales. As the pre| 
sures usually vary most, near the ends of the diagram a closer approximation 
can be made by subdividing the end strips, as is done in Fig. 20, which repri 
sents two steam engine indicator cards taken from opposite ends of the sarr 
cylinder and superimposed. The two diagrams are divided into ten equ; 
spaces and then each end space is subdivided. The mean heights of the sii 
divisions are measured and averaged to get the mean height of the whole en 
division, or average pressure in this case for the division. The average heigh 
of divisions for diagram No. 1 are set down in a column on the left, while tho; 




10 



30 40 50 GO 70 

Displacement in Per Cent of Stroke 



80 



yo 



100 



FIG. 20. Simpson's Method for Finding Mean Effective Pressure of Indicator Cards. 

for No. 2 are on the right; the sum of each column divided by ten and multiple 
by the spring scale gives the whole m.e.p. The heights of No. 1 in inch 
marked off continuously on a slip of paper measured a total of 11.16 ins. ai 
for No. 2, 10.79 ins., these quantities divided by 10 (number of strips) and muli 
plied by the spring scale, 50, gives the m.e.p., as before. This method 
often designated as Simpson's rule. 

The best and most commonly used method of area evaluation, whether f 
work or m.e.p. determination, is the planimeter, a well-known instrunie 
specially designed for direct measurements of area. 

14. Indicated Horse-power. Work done by the fluid in a cylinder, because 
is most often determined by indicator card measurements, measures the indicat 
horses-power, but the term is also applicable to work that would be done by t 
execution of a certain cycle of pressure volume changes carried out at a specifi- 
rate. The mean effective pressure in pounds per square inch, whether of ; 
indicator card or PV cycle, when multiplied by piston area in square inchc 



WORK AND POWER 45 

ives the average force acting on the piston for one stroke, whether the cycle 
squired one, two or x strokes for its execution, and this mean force multiplied 
y the stroke in feet gives the foot-pounds of work done by the cycle. Therefore, 

m.e.p. = mean effective pressure in pounds per square inch for the cycle 

referred to one stroke; 

a = effective area of piston in square inches; 
L = length of stroke in feet; 

n = number of equal cycles completed per minute; 
N = number of revolutions per minute; 
$ = mean piston speed = 2LN feet per minute; 

N 

z = number of revolutions to complete one cycle = . Then will the 

licated horse-power be given by, 

I.H.P.= 



33000 

(m.e.p. )LaN 
33000z 

(m.e.p. )aS 
33000X22* 



(6) 



(17) 



When there are many working chambers, whether in opposite ends of the 
cylinder or in separate cylinders, the indicated horse-power of each should 
e found and the sum taken for that of the machine. This is important not 
nly because the effective areas are often unequal, as, for example, in opposite 
&ds of a double-acting cylinder with a piston rod passing through one side 
inly or with two piston rods or one piston rod and one tail rod of unequal 
iameters, but also because unequal valve settings which are most common 
[ill cause different pressure volume changes in the various chambers. 

It is frequently useful to find the horse-power per pound mean effective pressure, 
r hich may be symbolized by K e , and its value given by 

_ Lan LaN 
~ 33000 = "330002" 

ping this constant, which may be tabulated for various values of n, stroke 
nd bore, the indicated horse-power is given by two factors, one involving 
pflinder dimensions and cyclic speed or machine characteristics, and the other 
jae resultant PV characteristic, of the fluid, symbolically, 

I.H.P.=12Xm.e.p.). 

foese tables of horse-power per pound m.e.p. are usually based on piston speed 
than rate of completion of cycles and are, therefore, directly applicable 



46 ENGINEERING THERMODYNAMICS 

when z = \ or n = 2N, which means that the two cycles are completed in or 
revolution, in which case, 



and 

aS 






33000' 
whence 

T TT -r TS- f \ (m.e.p.)a$ 

I.H.P.=g e (m.e.p.) P 



Table XI at the end of this chapter gives values of (H.P. per Ib. m.e.p.] 
K e for tabulated diameters of piston in inches and piston speeds in feet p< 
minute. Tables are frequently given for what is called the engine constan 
which is variously defined as either 

(a) oonrwv which must be multiplied by m.e.p. Xn to obtain H.P., or 
ooUUU 

(6) Qon , which must be multiplied by m.e.p. XLXn to obtain H.P. 
ooUUU 

For an engine which completes two cycles per revolution, this is the same j 
multiplying .by m.e.p. X$. Before using such a table of engine constants \ 
must be known whether it is computed as in (a) or in (6). 

Example. A 9 in. Xl2 in. double-acting steam engine runs at 250 R.P.M. and 1 
mean effective pressure is 30 Ibs. What is H.P. per pound m.e.p. and the I.H.P.? 

Lan 1X63.6X500 

Ke = 



33000 33000 

I.H.P.=. 9636X30 =28.908. 

Prob. 1. A pump has a piston speed of 250 ft. per minute; piston diameter is 24 ;k, 
What is the H.P. per pound mean effective pressure? 

Prob. 2. A simple single-acting 2-cylinder engine has a piston 10 ins. in diameter wji 
a 2-in. rod and a stroke of 15 ins. It runs with a mean effective pressure of 45 ;i 
per square inch at a speed of 220 R.P.M. What is the H.P.? 

Prob. 3. A gas engine has one working stroke in every four. If the speed is W 
R.P.M. what must be the m.e.p. to give 6 H.P. when the cylinder has a diameter of 6 s. 
and a stroke of 12 ins.? 

Prob. 4. An air compressor is found to have a mean effective pressure of 50 Ibs. u 
the cylinder is double acting and is 12 ins. diameter and 16 ins. stroke, what H.P. |B1 
be needed to drive it at 80 R.P.M.? 

Prob. 5. A gasoline engine has an engine constant (a) of .3. What must Ix 1 
m.e.p. to give 25 H.P.? 

Prob. 6. A blowing engine has an m.e.p. of 10 Ibs. Its horse-power is 500. VWj 
is the H.P. per pound m.e.p.? 



WORK AND POWER 47 

Prob. 7. Two engines of the same size and speed are so run that one gives twice the 
3ower of the other. How will the engine constants and m.e.p. vary? 

Prob. 8. From the diagrams following Section 9 what must have been the H.P. per 
pound m.e.p. to give 300 H.P. in each case? 

Prob. 9. How will the H.P. per pound m.e.p. vary in two engines if the speed of 
me is twice that of the other, if the stroke is twice, if the diameter of piston is twice? 

15. Effective Horse-power, Brake Horse-power, Friction Horse-power, 
Mechanical Efficiency, Efficiency of Transmission, Thermal Efficiency. Work 
s done and power developed primarily in the power cylinder of engines, and is 
transmitted through the mechanism with friction loss to some point at which 
t is utilized. There is frequently a whole train of transmission which may 
nvolve transformation of the energy into other forms, but always with some 
xosses, including the mechanical friction. For example, a steam cylinder may 
Irive the engine mechanism which in turn drives a dynamo, which transforms 
nechanical into electrical energy and this is transmitted to a distance over 
,vires and used in motors to hoist a cage in a mine or to drive electric cars. 
There is mechanical work done at the end of the system and at a certain rate, 
50 that there will be a certain useful or effective horse-power output for the 
system, which may be compared to the horse-power primarily developed in 
:he power cylinders. A similar comparison may be made between the primary 
)ower or input and the power left after deducting losses to any intermediate 
Doint in the system. For example, the electrical energy per minute delivered 
X) the motor, or motor input, is, of course, the output of the transmission line, 
^gain, the electrical energy delivered to the line, or electrical transmission 
nput, is the same as dj^namo output, and mechanical energy delivered to the 
lynamo is identical with engine output. The comparison of these measure- 
nents of power usually takes one of two forms, and frequently both; first, 
i comparison by differences, and second, a comparison by ratios. The ratio 
I )f any horse-power measurement in the system to the I.H.P. of the power cylin- 
i ler is the efficiency of the power system up to that point, the difference between 
.he two is the horse-power loss up to that point. It should be noted that, 
is both the dynamo and motor transform energy from mechanical to electrical 
>r vice versa, the engine mechanism transmits mechanical energy and the 
vires electrical energy, the system is made up of parts which have the 
unction of (a) transmission without change of form, and (7>) transformation 
>f form. The ratio of output to input is always an efficiency, so that the efficiency 
>/ the power system is the product of all the efficiencies of transformation and of 
ransfer or transmission, and the power loss of the system is the sum of trans- 
ormation and transmission losses. Some of these efficiencies and losses have 
eceived names which are generally accepted and the meaning of which is gen- 
Tally understood by all, but it is equally important to note that others have no 
lames, simply because there are not names enough to go around. In dealing 
vith efficiencies and power losses that have accepted names these names may 
vith reason be used, but in other cases where names are differently under- 
stood in different places or where there is no name, accurate description must be 



48 ENGINEERING THERMODYNAMICS 

relied on. As a matter of fact controversy should be avoided by definition 
of the quantity considered, whether descriptive names be used or not. 

Effective horse-power is a general term applied to the output of a machine, 
or power system, determined by the form of energy output. Thus, for an engine 
it is the power that might be absorbed by a friction brake applied to the shaft, 
and in this case is universally called Brake Horse-power. The difference between 
brake and indicated horse-power of engines is the friction horse-power of the 
engine and the ratio of brake to indicated horse-power is the mechanical efficiency 
of the engine. For an engine, then, the effective horse-power or useful horse- 
power is the brake horse-power. When the power cylinders drive in one machine 
a pump or an air compressor, the friction horse-power of the machine is the 
difference between the indicated horse-power of the power cylinders and that 
for the pump or compressor cylinders, and the mechanical efficiency is the 
ratio of pump or air cylinder indicated porse-power to indicated horse-power 
of the power cylinders. Whether the indicated horse-power of the air or pump 
cylinders can be considered a measure of useful output or not is a matter of 
difference of opinion. From one point of view the machine may be as considered 
built for doing work on water or on air, in which case these horse-powers may 
properly be considered as useful output. On the other hand, the power pump! 
is more often considered as a machine for moving water, in which case the 
useful work is the product of the weight of water moved into its head in feet, 
and includes all friction through ports, passages and perhaps even in pipes or 
conduits, which the indicated horse-power of the pump cylinder does not include, 
especially when leakage or other causes combine to make the pump cylinder 
displacement differ from the volume of water actually moved. With compressors 
the situation is still more complicated, as the air compressor may be considered 
useful only when its discharged compressed air has performed work in a rock 
drill, hoist or other form of an engine, in which case all sorts of measures of use- 
ful output of the compressor may be devised, even, for example, as the purely 
hypothetical^ possible work derivable from the subsequent admission and 
complete expansion of the compressed air in a separate air engine cylinder. 

Too accurate a definition, then, of output and input energy in machines anc 
power systems is not possible for avoidance of misunderstanding, which maj. 
affect questions both of power losses and efficiency of transmission and trans- 
formation whether in a power system or single machine. It is interesting t( 
note here that not only is the indicated work of the power cylinder always con-j 
sidered the measure of power input for the system or machine, but, as in thtj 
other cases, it is itself an output or result of the action of heat on the vapor o:[ 
gas and of the cycle of operations carried out. The ratio of the indicated powe [ 
or cylinder work, to the heat energy both in foot-pound units, that was expended 
on the fluid is the thermal efficiency of the engine referred to indicated horse j 
power or the efficiency of heat transformation into work, the analysis of whicl 
forms the bulk of the subject matter of Chapter VI. Similarly, the ratio o| 
any power measurement in the system to the equivalent of the heat supplied 
is the thermal efficiency of so much of the system as is included. 



WORK AND POWER 49 

Example. It has been found that when the indicated horse-power of an engine is 
>0, a generator is giving out 700 amperes at 220 volts. At end of a transmis ion line is a 
otor using the output of the generator." This motor on test gave out 180 brake horse- 
3wer. Assuming no losses in the transmission line, what was the efficiency of the 
.otor, of the generator, of the engine, and of the system? 

Motor efficiency = 



746 
NOTE: Volts X amperes = watts, and, watts -f- 746 =H. P. 

Engine and generator efficiency = =82.4%. 

74o 

250 

Efficiency of system = =72% or 82.4x87.2 =72%. 
Zoo 

b. 1. An engine is belted to a pump; the I.H.P. of the engine is 50, of the pump 
', and the pump delivers 1200 gallons water per minute against 100-ft. head. What is 
he efficiency of each part and of the entire system. 

Prob. 2. An engine is geared to air compressor. Upon test, the efficiency of the engine 
lone, gearing alone and compressor alone were each 80 per cent. When the com- 
ressor H.P. was 100 what was that of the engine? 

Prob. 3- A water-wheel is run by the discharge from a pump. The B.H.P. of wheel 
j found to be 20 when the pump is delivering 45 gallons of water per minute at a 
.lead of 1000 Ibs. per square inch. The water I.H.P. of the pump is 30 and the 
(team I.H.P. is 40. What are the efficiencies of each part of the system and the over-all 
fficienc} 7 '? 

Prob. 4. Perry gives a rule for the brake horse-power of steam engines as being 
iqual to .95 I.H.P. 10. On this basis find the mechanical efficiency of a 500-H.P. 
jngine from 200 to 500 H.P. Show results by a curve with-B.H.P. and per cent efficiency 
U coordinates. 

Prob. 5. Perry gives a rule for the efficiency of an hydraulic line as H = .71 25 where 
tf is the useful power of the pump and / is the indicated. Find / for values of H from 
1 00 to 300 and plot a curve of results. 

Prob. 6. An engine gives one I.H.P. for every 3 Ibs. of coal per hour. One pound 
|f coal contains 9,500,000 ft.-lbs. of energy. What is the thermal efficiency? 

Prob. 7. A gas engine has a mechanical efficiency of 70 per cent when delivering 
tower to a generator which in turn has an efficiency of 90 per cent. If the engine uses 
j.5 cu.ft. of gas per indicated horse-power hour and the gas contains 700,000 ft.-lbs. per 
; 'iibic foot, what is the net thermal efficiency of the system? 

16. Specific Displacement, Quantity of Fluid per Hour or per Minute per 

..H.P. It has been shown that the work done in cylinders by pressure volume 
Changes of the vapor or gas depends on the mean effective pressure and on the 
jlisplacement, or that there is a relation between I.H.P. and displacement. 
The quantity of fluid used also depends on the displacement and may be expressed 






50 ENGINEERING THERMODYNAMICS 

in cubic feet per minute at either the low pressure or high pressure con- 
dition when the work is done between two definite pressure limits, or in terau 
of pounds per minute or hour, which involves the application of fluid densities 
to volumes and which eliminates the double expression for the two conditions 
of pressure. The displacement per hour per horse-power, termed the specific 
displacement, is the basis of computations on the steam consumption of steair 
engines, the horse-power per cubic feet of free air per minute for air compressors 
the horse-power per ton refrigeration for refrigerating machines and the con- 
sumption of fuel per hour per horse-power for gas and oil engines. It is 
therefore, a quantity of great importance in view of these applications. Apply- 
ing the symbols already defined to displacement in one direction of one side oJ 
a piston 

Displacement in cu.ft. per stroke = I/X 



144 ' 
Displacement in cu.ft. per minute = LX 



Displacement in cu.ft. per hour = 



T ,. 

Indicated horse-power 



(m.e.p.)LaN 



330QO 330 oOg 



Whence expressing displacement per hour per I.H.P. or specific displacemen 
in one direction for one side of a piston by D S) 

1375Qg 



^ ^ = 

(m.e.p.)LaN 144(m.e.p.) (m.e.p.)* 
330002 






From Eq. (19) it appears that the specific displacement is equal to zX!3,75( 
divided by the mean effective pressure in pounds per square inch. 

If two points, A and B, be so located on the indicator card, Fig. 21, as t< 
have included between them a fluid transfer phase, either admission to, o 
expulsion from the cylinder, then calling B a = pounds per cubic foot or densit; 
at point A, and B 6 = pounds per cubic foot or density at point B } the weigh 
of fluid present at A is, 

(D a +Ct)8 a Ibs., 
and weight of fluid present at B is 

(Db-^-Cl)^b Ibs., 

whence the weight that has changed places or passed in and out per stroke ifj 
(Di-\-Ct)ou (D a -i-Ci)da Ibs. per stroke. 



WORK AND POWER 



51 



both A and B lie on the same horizontal as A and B', d a = d b = d, the 
^nsity of fluid at the pressure of measurement, whence the weight of fluid 
n sed per stroke, will be 



ad the volume per stroke used at density d is 

Db'-Da CU.ft., 

rhich compared to the displacement is 

Db'-Da 

D ' 



\A 



\A 



B' 



21. Determination of Consumption of Fluid per Hour per Indicated Horse-power from 

the Indicator Card. 

Phis is the fraction of the displacement representing the volume of fluid pass- 
ng through the machine at the selected pressure. Multiplying the specific 
lisplacement by this, there results, 

Cu.ft. of fluid per hr. at density (d) per I.H.P.= ^ 5 Db '~ Da , 



md 



Lbs. of fluid per hr. per I.H.P = 



(20) 



ore generally, that is, when A and B are not taken at the same pressures 
Lbs. fluid per hr. per I.H.P. = ^ 3 J 5 \(D b +Cl)d l> - (D+C7)$.~| . (21) 



The particular forms which this may take when applied to special cases will be 
examined in the succeeding chapters. 



52 ENGINEERING THERMODYNAMICS 

Example. An air compressor whose cylinder is 18x24 ins. (18 ins. in diameter and 
stroke 2 ft.) runs at 60 R.P.M. and is double acting. The mean effective air pressure 
is 50 Ibs. per square inch. What is the specific displacement? 



Cu.ft. per hour = ~p = 60 X2 x X120 =25,600. 



T IT P _m.e.p.Layi _50x2x254.5xl20 _ Q9 
: ~33,000 33,000 



Cu.ft. per hour = 25,600 = 
I.H.P. = 92.3 = 



or by the formula directly, 



13,750 13,750 0>7 , 
= -=^/o. 
m.e.p. 50 

Prob. 1. What will be the cubic feet of free air per hour per horse-power delivered 
by a 56x72-in. blowing engine with 4 per cent clearance and mean effective pressure of 
10 Ibs. per square inch? 

Prob. 2. An 18x22-in. ammonia compressor works with a mean effective pressure 
of 45 Ibs. per square inch. What is the weight of NH S per I.H.P. hour if the speed is 
50 R.P.M. and compressor is double acting having a volumetric efficiency of 90 per 
cent? Use tabular NH 3 densities. 

Prob. 3. A steam engine whose cylinder is 9x12 ins. runs at a speed of 300 R.P.M. 
and is double acting. If the m.e.p. is 60 Ibs. and the density of steam at end of the 
stroke is .03, how many pounds of steam are used per hour per indicated horse-power? 

17. Velocity Due to Free Expansion by PV Method. All the cases examined 
.tor the work done by PV cycles have been so far applied only to their execution 
in cylinders, but the work may be developed in nozzles accelerating the gas 
or vapor in free expansion, giving, as a consequence, a high velocity to the fluid. 
It was noted that for cylinders many combinations of phases might be found 
worthy of consideration as typical of possible actual conditions of practice, 
but this is not true of proper nozzle expansion, which has but one cycle, that of 
Fig. 22. That this is the cycle in question is seen from the following considera- 
tions. Consider a definite quantity of the gas or vapor approaching the nozzle 
from a source of supply which is capable of maintaining the pressure. It pushes 
forward that in front of it and work will be done, ABCD, equal to the admission 
of the same substance to a cylinder, so that its approach A B may be considered 
as a constant pressure, volume increasing phase for which the energy comes 
from the source of supply. This same substance expanding to the lower pres- 
sure will do the work CBEF; but there will be negative work equivalent to the 
pushing away or displacing of an equivalent quantity of fluid at the low pres- 
sure, or FEGD, making the work cycle A BEG, in which AG is the excess of 



WORK AND POWER 53 

nitial over back pressure or the effective working pressure, remaining constant 
luring approach and lessening regularly during expansion to zero excess at E. 
The work done will be from Eq. (13), 



D 



c 



|IG. 22. Pressure- Volume Diagram for Noz/le Expansion Measuring the Acceleration 
Velocity and Horse-power of Jets. 



s-l 



s-l 






Whence 



r 1 



54 ENGINEERING THERMODYNAMICS 

Assuming the initial velocity to be zero, and the work of Eq. (22) to be don 
on 1 lb., the final or resultant velocity will be according to Eq. (6), 

u = V%iW 



or 

(24 



This velocity is in feet per second when pressures are in pounds per square foo 
and volumes in cubic feet of 1 lb. of substance, and is known as Zeuner 
equation for the velocity of a gas or vapor expanding in a nozzle. It i 
generally assumed that such expansion, involving as it does very rapid motio 
of the fluid past the nozzle, is of the adiabatic sort, as there seems to be n 
time for heat exchange between fluid and walls. As already noted, the vain 
of s for adiabatic expansion of vapors is not constant, making the correc 
solution of problems on vapor flow through orifices practically impossible b 
this method of pressure volume analysis, but as will be seen later the therm* 
method of solution is exact and comparatively easy. 

NOTE. A comparison of Eqs. (22) and (13) and the figures correspondi 
will show that the area under the process curve, which is the same as the wo 
done during the compression or expansion, if multiplied by s will equal the ar 
to the left of the process curve, which in turn represents, as in Fig. 21, 
engines, the algebraic sum of admission, complete expansion, and exhaust wo 
areas, or as in Fig. 9 for compressors, the algebraic sum of suction, compressi 
and delivery work areas. This statement must not be thought to refer to t 
work area of such a cycle as that of Fig. 8, where expansion is incomplete, nor 
case of over-expansion, Fig. 10. 

Example. In Fig. 22 assume the initial pressure at 100 Ibs. per square inch absolu 
back pressure at atmosphere, and expansion as being adiabatic. What will be 
work per pound of steam and the velocity of the jet, if Vb is 4.36 cu.ft. and s = 1.3 
superheated steam? 



= 27,206 X. 608 = 16,541 ft.-lbs.; 



= 1028 ft. per second. 



WORK AND POWER 



55 



Prob, 1. Taking the same pressure range as above, find W and u for adiabatic expan- 
sion of air, also for isothermal expansion. 

Prob. 2. How large must the effective opening of the suction valve be, in an air 
compressor 18x24 ins. to allow the cylinder to properly fill if the mean pressure-drop 
through the valve is 1 Ib. per square inch and the compressor runs at 80 R.P.M.? 

Prob. 3. What must be the diameter of the inlet valve in a gas engine to fill a 
cylinder of i cu.ft. capacity if the lift of the valve is f ins., allowing a pressure drop of 
1 Ib. per square inch? Engine makes 150 working strokes per minute. 

Prob. 4. It has been found from experiment that the velocity of air issuing from a hole 
jin plate orifice is 72 per cent of what would be expected from calculation as above when 
ithe absolute pressure ratio is 2 to 1, and 65 per cent when the absolute pressure ratio 
jis 1^ to 1. What will be the actual velocity for air flowing from a tank to atmosphere 
|for these pressure ratios? 

Prob. 5. CO 2 stored in a tank is allowed to escape through an orifice into the air. 
What will be the maximum velocity of the jet if the pressure on the tank be 100 Ibs. 
per square inch gage? 

NOTE: 1 Ib. C0 2 at pressure of 100 Ibs. per square inch gage occupies 1.15 cu.ft. 

Prob. 6. If ammonia gas and hydrogen were allowed to expand from the same pres- 
ire, how would their maximum velocities compare? Vol. of 1 Ib. of NH 3 at 50 Ibs. per 
;square inch gage is 4.5 cu.ft. Vol. of 1 Ib. of H at same pressure is 77.5 cu.ft. 

18. Weight of Flow through Nozzles. Applying an area factor to the 
velocity equation will give an expression for cubic flow per second which 
I becomes weight per second by introducing the factor, density. 

Let the area of an orifice at the point of maximum velocity, u, be A sq.ft., 
then will the cubic feet per second efflux be Au. Assume the point of maxi- 
mum velocity, having area A, to be that part of the nozzle where the pressure 
ihas fallen to P e , Fig. 22, and the gas or vapor to have the density B e pounds 
per cubic foot. Then will the nozzle flow in pounds per second be 



ut the weight per cubic foot is the reciprocal of the cubic feet per pound, 
' e , which it has already been assumed, is the final volume, of one pound of the 
| fluid. Hence, 

uA 



w 



TV 



This may be put in terms of initial gas or vapor conditions for, 



V T7\ I 
V c V b\ ^r 



Tience 



tt 



uA 



uA 



56 ENGINEERING THERMODYNAMICS 

Substituting in this the value of u from Eq. (24), 



""- | - c - (25) 

This weight will be a maximum for a certain value of the pressure ratio, depend- 
ing on the value of s only, and this value can be found by placing the first dif- 

ferential coefficient of w with respect to ( ^ ) equal to zero. 



dw o. 



To accomplish this, rearrange Eq: (25) as follows: 

2 8+1 



But as the other factors do not enter to effect the result so long as P& does not 
vary, w is a maximum when the bracket 



s + 1 



is a maximum or when 

2 



or 



or 



_ 
But as ( p^) S cannot be equal to zero in practice, then 



which gives the condition that w is a maximum when 



/PeVV _S + 1 

\Pb) 2 ' 



WORK AND POWER 57 

r maximum flow for given initial pressure occurs when 



)' ' (26) 



For air expanding adiabatically s= 1.407. Maximum flow occurs when 
) 
- = .528 and for most common values of s it will be between .50 and .60 

*6 

j?his result is quite remarkable and is verified by experiment reasonably closely. 
It shows that, contrary to expectation, the weight of efflux from nozzles will not 
Continuously and regularly increase with increasing differences in pressure, but 
lor a given initial pressure the weight discharged per second will have reached 
Us limit when the final pressure has been diminished to a certain fraction of the 
Initial, and any further decrease of the discharge pressure will not increase the 
{low through an orifice of a given area. 

The subject of flow in nozzles will be treated more completely in 
jvhapter VI. 

Prob. 1. For the following substances under adiabatic expansion determine the 
jressure ratio for maximum flow and find the rate of flow per square inch of orifice under 
[his condition when flow is into a vacuum of 10 ins. of mercury with standard barometer: 

(a) Carbon dioxide. 

(6) Nitrogen. 

(c) Hydrogen. 

(d) Ammonia. 

(e) Dry steam according to saturation law. 

19. Horse-power of Nozzles and Jets. Although, strictly speaking, nozzles 
[an have no horse-power, the term is applied to the nozzle containing the 
i.rifice through which flow occurs and in which a certain amount of work is 
j.one per minute in giving to a jet of gas or vapor initially at rest a certain 
linal velocity, and amount of kinetic energy. The foot-pounds of work per 
Sound of fluid multiplied by the pounds flowing per second will give the foot- 
pounds of work developed per second within the nozzle, and this divided by 
';50 will give the horse-power developed by the jet, or the nozzle horse-power, 
accordingly, 




-^X^gVx. (a) 

(6) 



8-1 , 3 



(27) 



58 ENGINEERING THERMODYNAMICS 






where the expression in the bracket is the work done per pound of substance 
The pressures are expressed in pounds per square foot, areas in square feei 
and volumes in cubic feet. 

Example. A steam turbine operates on wet steam at 100 Ibs. per square inch abso 
lute pressure which is expanded adiabatically to atmospheric pressure. What must b< 
the area of the nozzles if the turbine is to develop 50 H.P. ideally? 

NOTE : 1 cu.ft. of steam at 100 Ibs. = .23 Ib. 

_ s_ 

By Eq. (26), maximum flow occurs when the pressure ratio is ( ) S , or, fo 



1.11 
this case when the pressure is 100 -s- (^~rr) =58 Ibs. per square inch absolute. A; 

the back pressure is one atmosphere, the flow will not be greater than for the abov 
critical pressure. Substituting it in Eq. (25) will give the flow weight w, and using th< 
actual back pressure in Eq. (22) will give the work W. 

.11 



= 110000 ft.-lbs. per pound of steam. 



= 198A Ibs. of steam per second. 
By Eq. (27 a), 

^WXffl _ 110000 X198A 

550 = 550 

Whence 

50 



Prob. 1. What will be the horse-power per square inch of nozzle for a turbine usin 
hot gases if expansion follows law PVs =k, when s = 1.37, the gases being at a pressure c 
200 Ibs. per square inch absolute and expanding to atmosphere. 

Let the volume per pound at the high pressure be 2 cu.ft. 

Prob. 2. What will be the horse-power per square inch of nozzle for the problem 
of Section 17? 

Prob. 3. Suppose steam to expand according tolawPFs=/b, where s = 1.111, fror 
atmosphere to a pressure of 2 Ibs. per sq. inch absolute. How will the area of the or 
fice compare with that of the example to give the same horse-power? 

NOTE: V b =2QA. 

Prob. 4. Suppose steam to be superheated in the case of the example and of the la* 
problem, how will this affect the area of nozzle? 

NOTE: Let V* =5 and 32 respectively. 

Prob. 5. How much work is done per inch of orifice if initial pressure is 100 Ib; 
absolute on one side and final 10 Ibs. absolute on other side of a valve through wbic 
air is escaping? 






WORK AND POWER 59 



GENERAL PROBLEMS ON CHAPTER I 



1. An air compressor is required to compress 500 cu.ft. of free air per minute to a 
pressure of 100 Ibs. per square inch gage; the compressor is direct connected to a steam 
engine. The mechanical efficiency of the machine is 80 per cent. What will be the 
steam horse-power if compression is (a) isothermal; (b) adiabatic? 

2. A mine hoist weighing 10 tons is raised 4500 ft. in one minute. In the first 20 
seconds it is accelerated from rest to a speed of 6000 ft. per minute; during the next 30 
seconds speed is constant at this value, and during last 10 seconds it is brought to rest. 
What will be (a) work of acceleration for each period; (6) work of lift for each period; 
(c) total work supplied*by engine; (d) horse-power during constant velocity period? 

3. The engine driving the above hoist is driven by compressed air. If air is supplied at 
a pressure of 150 Ibs. per square inch gage and is admitted for three-quarters of the 
'stroke, then expanded adiabatically for the remainder of the stroke and exhausted to the 
atmosphere find (a) what must the piston displacement be to lift the hoist, the work of 
acceleration being neglected? (6) To what value could the air pressure be reduced if 
air were admitted full stroke? 

4. It is proposed to substitute an electric motor for the above engine, installing a 
water-power electric plant at a considerable distance. The type of wheel chosen is one in 
which a jet of water issuing from a nozzle strikes against a series of revolving buckets. 
The available head at the nozzle is 1000 ft. Assuming the efficiency of the motor to be 
85 per cent, transmission 80 per cent, generators 90 per cent, and water-wheels 60 per 
cent, what will be the cubic feet of water per minute? 

5. A steam turbine consists of a series of moving vanes upon which steam jets issuing 
from nozzles impinge. It is assumed that for best results the speed of the vanes should be 
half that of the jets. The steam expands from 100 Ibs. per square inch gage to 5 Ibs. 
per square inch absolute, (a) What must be the best speed of vanes for wet steam where 

s = l.lll? (6) If 55 per cent of the work in steam is delivered by the wheel what must 
I be the area of nozzle per 100 H.P., and weight of steam per hour 100 H.P? 

NOTE: F 6 =3.82. 

6. It has been found that a trolley car uses a current of 45 amperes at 550 volts when 
li running 12 miles per hour. If the motor is 80 per cent efficient, what is tractive effort? 

NOTE : Volts X amperes = watts, and watts -f- .746 = H.P. 

7. An engine whose cylinder is 18x24 ins. (diameter = 18 ins., stroke =24 ins.) is 
double acting and runs at a speed of 125 R.P.M. Steam is admitted for one-quarter 
stroke at a pressure of 125 Ibs. per square inch gage, allowed to expand for the rest of 

:i the stroke and then exhausted into a vacuum of 28.25 ins. of Hg. (a) Draw a PV 
diagram of the cycle and find the m.e.p. from the diagram and by calculation, and then 
find the horse-power. (6) Consider steam to be admitted one-half stroke without 
other change. How will the horse-power vary? (c) What will be the horse-power for 
one-quarter admission if the exhaust pressure is 15 Ibs. per square inch absolute? (d) 
What will be the horse-power if the steam pressure be made 150 Ibs. per square inch 
absolute, other conditions as in (a)? (e) Suppose all conditions as in (a) but the 

speed lowered to 75 R.P.M. What will be the horse-power? 

8. Assuming that 50 per cent of the work in the jet is transformed to useful work, 
what must be the total area of the nozzles of a steam turbine to develop the same horse- 



60 ENGINEERING THERMODYNAMICS 

power as the engine in problem (7 a), the pressure range being the same and s being 1.3? 
7^=3.18. 

9. Water is being pumped from the bottom of a shaft 700 ft. deep at the rate of 1000 
gallons per minute by an electrically driven pump. Efficiency of pump is 70 per cent, 
motor 90 per cent, transmission line 95 per cent, generator 85 per cent, and mechanical 
efficiency of engine 80 per cent. What will be the indicated horse-power of the engine? 
If the above installation were replaced with an air-driven pump of 65 per cent efficiency, i 
efficiency of transmission being 100 per cent, and that of the compressor and engine: 
80 per cent, what would be the horse-power of this engine? 

10. Show by a PV diagram, assuming any convenient scales, that the quantity of air 
discharged byacompressor and the horse-power, both decrease as the altitude increases, and 
that the horse-power per cubic foot of air delivered increases under the same condition. 

11. A centrifugal pump is driven by a steam engine directly connected to it. The 1 
pump is forcing 1000 gallons of water per minute against a head 'of 250 ft. and runs at a 
speed of 450 R.P.M. The engine is double acting and its stroke equals the diameter of 
the cylinder. Steam of 100 Ibs. per square inch gage is admitted for half stroke, allowed 
to expand the rest of the stroke so that s = 1, and is then exhausted to atmosphere. Whatj 
must be the size of the engine if the pump efficiency is 65 per cent and the engine 
efficiency 75 per cent? 

12. (a) What will be the pounds of steam used by this engine per hour per hoi 
power? (6) If the steam were admitted but one-quarter of the stroke and the initialj 
pressure raised sufficiently to maintain the same horse-power, what would be the new 
initial pressure and the new value of the steam used per horse-power per hour? 

NOTE: Weight of steam per cubic foot for (a) is .261; for (6) is .365. 

13. If it were possible to procure a condenser for the above engine so that the exhaust 
pressure could be reduced to 2 Ibs. per square inch absolute, (a) how much would the 
power be increased for each of the two initial pressures already given? (6) How would 
the steam consumption change? 

14. A motor-fire engine requires a tractive force of 1300 Ibs. to drive it 30 miles perl 
'hour, its rated speed. The efficiency of engine and transmission is 80 per cent. When 
the same engine is used to actuate the pumps 70 per cent of its power is expended OD 
the water. What will be the rating of the engine in gallons per minute when pumping 
against a pressure of 200 Ibs. per square inch? 

15. A compressor when compressing air at sea level from atmosphere to 100 Ibs. pel 
square inch absolute, expends work on the air at the rate of 200 H.P., the air being com- 
pressed adiabatically. (a) How many cubic feet of free air are being taken into the 
compressor per minute and how many cubic feet of high pressure air discharged? 
Compressor is moved to altitude of 8000 ft. (6) What will be the horse-power if the same 
amount of air is taken in and how many cubic feet per minute will be discharged? 

(c) What will be the horse-power if the same number of cubic feet are discharged as in 
case (a) and what will be the number of cubic feet of low pressure air drawn in? 

(d) Should superheated ammonia be substituted for air at sea level, what would be the 
necessary horse-power? 

16. A car weighing 40 short tons is at rest and is struck by a train of four cars, each 
of the same weight as the first, (a) Upon impact the single car is coupled to the train and 
all move off at a certain velocity. If the original velocity of the train was 3 miles 
per hour, what will it be after attachment of the extra car? (b) If instead of coupling, 
the extra car after impact moved away from the train at twice the speed the train was 
then moving, what would be the speed of train? 



WORK AND POWER , 61 

17. To drag a block of stone along the ground requires a pull of 1000 Ibs. If it be 
placed on rollers the pull will be reduced to 300 Ibs., while if it be placed on a wagon with 
veil-made wheels, the pull will be but 200 Ibs. Show by diagram how the work required 
;o move it 1000 ft. will vary. 

18. 100 cu. ft. of air, at atmospheric pressure, is compressed to six times its 
)riginal pressure, (a) What will be the difference in horse-power to do this in 45 seconds 
sothermally and adiabatically at an elevation of 8000 ft. (6) What will be the final vol- 
imes? (c) What will be the difference in horse-power at sea level? (d) What will be the 
inal volumes? 

19. An engine operating a hoist is run by compressed air at 80 Ibs. per square inch 
;age. The air is admitted half stroke, then expanded for the rest of the stroke so that 
-'=1.3 and then exhausted to atmosphere. The engine must be powerful enough to lift 
i, ton cage and bring it up to a speed of 2000 ft. per minute in 15 seconds. What will be 
jhe necessary displacement per minute? 

20. Construct PV diagrams for Probs. 1, 11, 13 and 15, showing by them that the 
jyork of admission, compression or expansion, and discharge or exhaust, is equal to that 
jound algebraically. 

21 The elongation of wrought iron under a force F is equal to the force times the 
3ngth of the piece divided by 25,000,000 times the cross-section of metal in the piece. 
L4^ in. pipe has 3.75 sq. ins. of metal; a line of this pipe 1 mile long is running full of 
rater with a velocity of 600 ft. per minute. This is stopped in 1 second by closing a 
ralve. Assuming pipe did not burst, what would be the elongation? 

22. Two steam turbines having nozzles of equal throat areas are operating on a steam 
liressure of 150 Ibs. per square inch gage. One is allowing steam to expand to atmosphere 
Hie other to 2 Ibs. per square inch absolute, both cases having an exponent for expansion 
If 1.11. Find the relation of the horse-power in the two cases. 

23. The power from a hydro-electric plant is transmitted some distance and then 
ued to drive motors of various sizes. At the time of greatest demand for current it has 
seen found that 1000 horse-power is given out by the motors. Taking the average 
ijfficienoy of the motors as 70 per cent, transmission efficiency as 85 per cent, generator 
Ifficiency as 85 per cent, and water-wheel efficiency as 70 per cent, how many cubic feet 
If water per second will the plant require if the fall is 80 ft.? 

24. A small engine used for hoisting work is run by compressed air. Air is admitted 
|)r three-quarters of the stroke and then allowed to expand for the rest of the stroke in 

jch a way that s = 1 .4 and finally exhausted to atmosphere. For the first part of the 
oisting,full pressure (80 Ibs. per square inch gage) is applied, but after theload has been 
dcelerated the pressure is reduced to 30 Ibs.. per square inch gage. If the engine has 
#o cylinders each 9 Xl2 in., is double acting and runs at 200 R.P.M., what will be the 
i) horse-power in each case, (6) the specific displacement? 

\ 25. A steam engire has a cylinder 12x18 in., is double acting, and runs at 150 
['..P.M. (a) What is the engine constant, and (b) horse-power per pound m.e.p.? 

ft 26. A water-power site has available at all times 3500 cu.ft. of water per min- 

ffe at a 100-ft. fall. Turbines of 70 per cent efficiency are installed which take the place 
l two double-acting steam engines whose mechanical efficiencies were 85 per cent. 

fie speed of the engines was 150 R.P.M., m.e.p. 100 Ibs. per square inch, and stroke was 

esjvice the diameter. What was the size of each engine? 

i! 27. Assuming the frictional losses in a compressor to have been 15 per cent, how many 

jkf*ft. of gas per minute could a compressor operated by the above engines compress 
BI atmosphere to 80 Ibs. per square inch gage if s = 1 .35? 



62 



ENGINEERING THERMODYNAMICS 



TABLE I 
CONVERSION TABLE OF UNITS OF DISTANCE 



Meters. 1 


Kilometers. 


Inches. 


Feet. 


Statute Miles. 


Nautical Miles. 


1 


0.001 


39.37 


3.28083 


0.000621370 


0.000539587 


1000 


1 


39370.1 


3280.83 


0.62137 


0.539587 


0.0254 


0.0000254 


1 


0.083333 


0.0000157828 


0.0000137055 


0.304801 


0.0003048 


12 


1 


0.000189394 


0.000164466 


1609.35 


1.60935 


63360 


5280 


1. 


0.868382 


1853.27 


1.85327 


72963.2 


6080.27 


1.15157 


1. 



1 In accordance with U. S. Standards (see Smithsonian Tables). 

TABLE II 
CONVERSION TABLE OF UNITS OF SURFACE 



Sq. Meters. 


Sq. Inches. 


Sq. Feet. 


Sq. Yards. 


Acres. 


Sq. Miles. 


1 

.000645 
.0929 


1550.00 
1 
144 


10.76387 
.00694 
1 


1.19599 
.111 


.000247 




.8361 


1296 


9 


1 


.000206 




4046 87 




43560 


4840 


1 


001562 


2589999 




27878400 


3097600 


640 


1 















TABLE III 
CONVERSION TABLE OF UNITS OF VOLUME 



Cu. Meters. 


Cu. Inches. 


Cu. Feet. 


Cu. Yards. 


Lities 
(1000 Cu. Cm.) 


Gallons (U.S.) 


1 


61023.4 
1 


35.3145 
.000578 


1.3079 


1000 
.016387 


264.170 
00433 


.028317 
. 76456 


1728 
46656 


1 
27 


.03704 
1 


28.317 


7 . 4805 
201 974 


.001 
.003785 


61.023 
231 


.035314 
. 13368 


.001308 
.004951 


1 

3.7854 


.26417 

1 



TABLES 



63 



TABLE IV. 



CONVERSION TABLE OF UNITS OF WEIGHT AND FORCE 



Kilogrammes. 


Metric Tons. 


Pounds. 


U. S. or Short Tons. 


British or Long Tons. 


1. 


0.001 


2.20462 


0.00110231 


0.000984205 


1000. 


1. 


2204.62 


1 . 10231 


0.984205 


0.453593 


0.000453593 


1. 


0.0005 


0.000446429 


907.186 


0.907186 


2000. 


1. 


0.892957 


1016.05 


1.01605 


2240. 


1.12000 


1. 



TABLE V 
CONVERSION TABLE OF UNITS OF PRESSURE 





Pounds per 
Square Foot. 


Pounds per 
Square Inch. 


Inches of 
Mercury at 
32 F. 


Atmospheres 
(Standard at 
Sea Level). 


'ie Ib. per sq. ft 


1 


0.006944 


0.014139 


0004724 


le Ib per sq. in. . . 


144 


1 


2 03594 


06802 


'ie ounce per sq. in 
lie atmosphere (standard at sea 
level) 
lie kilogramme per square meter . . 
*ie gramme per square millimeter . 
tie kilogramme per square centi- 
meter 


9. 

2116.1 
20.4817 
204.817 

2048 17 


0.0625 

14.696 
0.142234 
1.42234 

14 2234 


0.127246 

29 . 924 
0.289579 
2.89579 

28 9579 


0.004252 

1. 
0.009678 
0.09678 

9678 


FLUID PRESSURES 

lie ft. of water at 39.1 F. (max. 
dens.) .... 


62 . 425 


0.43350 


0.88225 


029492 


lie ft. of water at 62 F 
l(ie in. of water at 62 F. . . 


62.355 
5.196 


0.43302 
. 036085 


0.88080 
. 07340 


0.029460 
002455 


ic in. of mercury at 32 F. (stand- 
lard) ' . . 


70 7290 


491174 


1. 


033416 


!>iie centimeter of mercury at C. . 
|ie ft. of air at 32 F., one atmos. 
press 


27.8461 
08071 


0.193376 
0005604 


0.393701 
0011412 


0.013158 
00003813 


lie ft. of air, 62 F 


0.07607 


0005282 


0010755 


00003594 













I * PRESSURES MEASURED BY THE MERCURY COLUMN. For temperatures other than 32 F., the density 
'mercury, pounds per cubic inch, and hence the pressure, pounds per square inch, due to a column of mercury 
ich high, is given with sufficient accuracy by the following formula: 

p = 0.4912-a-32) X0.0001. 

The mercurial barometer is commonly made with a brass scale which has its standard or correct length 
5 F, and a linear coefficient of expansion of about 0.000001 for each degree Fahrenheit. Hence, to 
<rect the standard of mercury at 32 F., the corrected reading will be 






'ere H t is the observed height at a temperature of t F. 



64 



ENGINEERING THERMODYNAMICS 



TABLE VI 
CONVERSION TABLE OF UNITS OF WORK 



Kilogrammeters. 


Foot-pounds. 


Foot Tons (Short Tons). 


Foot Tons (Long Tons). 


1. 


7.23300 


0.00361650 


0.00322902 


0.138255 


1. 


0.000500 


0.000446429 


276.510 


2000. 


1. 


0.892857 


309.691 


2240. 


1 . 12000 


1. 



See also more complete table of Units of Work and Energy in Chapter IV on Work and Heat. 

TABLE VII 
CONVERSION TABLE OF UNITS OF POWER 






Foot-pounds per 
Second. 


Foot-pounds per 
Minute. 


Horse-power. 


Cheval-Vapeur. 


Kilogrammeters pe 
Minute. 


1. 


60. 


0.00181818 


0.00184340 


8.29531 


0.0166667 


1. 


0.000030303 


0.0000307241 


. 138252 


550.000 


33000. 


1. 


1.01387 


4562.42 


542.475 


32548.5 


0.986319 


1. 


4500.00 


0.120550 


7.23327 


0.000219182 


0.000222222 


1. 



TABLE VIII 
UNITS OF VELOCITY 





Feet per Minute. 


Feet per Seconi 


One foot per second . ... 


60. 


1. 


One foot per minute . . 


1. 


0.016667 


One statute mile per hour 


88. 


1.4667 


One nautical mile per hour = 1 knot . . 


101.338 


1.6890 


One kilometer per hour 


54 6806 


0.911344 | 


One meter per minute 


3.28084 


0.054581 


One centimeter per second 


2.00848 


0.032808 









TABLES 



65 



TABLE IX 
TABLE OF BAROMETRIC HEIGHTS, ALTITUDES, AND PRESSURES 

(Adapted from Smithsonian Tables) 

Barometric heights are given in inches and millimeters of mercury at its standard density 
(32 F.). 

Altitudes are heights above mean sea level in feet, at which this barometric height is 
standard. (See Smithsonian Tables for corrections for latitude and temperature.) 

Pressures given are the equivalent of the barometric height in Ibs. per sq. in. and per 
sq. ft. 



Standard Barometer. 


Altitude, Feet above 
Sea Level. 


Pressure, Pounds per 


Inches. 


Centimeters. 


Square Inch. 


Square Foot. 


17.0 


43.18 


15379 


8.350 


1202.3 


17.2 


43.69 


15061 


8.448 


1216.6 


17.4 


44.20 


14746 


8.546 


1230.7 


17.6 


44.70 


14435 


8.645 


1244.8 


17.8 


45.21 


14128 


8.742 


1259.0 


18.0 


45.72 


13824 


8.840 


1273.2 


18 2 


46.23 


13523 


8.940 


1287.3 


18.4 


46.73 


13226 


9.038 


1301.4 


18.6 


47.24 


12931 


9.136 


1315.6 


18.8 


47.75 


12640 


9.234 


1329.7 


19.0 


48.26 


12352 


9.332 


1343.8 


19.2 


48.77 


12068 


9.430 


1357.9 


19 .4 


49.28 


11786 


9.529 


1372.1 


19.6 


49.78 


11507 


9.627 


1386.3 


19.8 


50.29 


11230 


9.726 


1400.4 


20.0 


50.80 


10957 


9.825 


1414.6 


20.2 


51.31 


10686 


9.922 


1428.7 


20 4 


51.82 


10418 


10.020 


1442.9 


20^6 


52.32 


10153 


10.118 


1457.0 


20.8 


52.83 


9890 


10.217 


1471.2 


21.0 


53.34 


9629 


10.315 


1485.3 


21 .2 


53.85 


9372 


10.414 


1499.4 


21 4 


54.36 


9116 


10.511 


1513.6 


21 .6 


54.87 


8863 


10.609 


1527.7 


21.8 


55.37 


8612 


10.707 


1541.8 


22.0 


55.88 


8364 


10.806 


1556.0 


22.2 


56.39 


8118 


10.904 


1570.1 


22.4 


56.90 


7874 


11.002 


1584.3 


22 6 


57.40 


7632 


11.100 


1598.4 


22.8 


57.91 


7392 


11.198 


1612.6 


23.0 


58.42 


7155 


11.297 


1626.7 


23.2 


58.92 


6919 


11.395 


1640.8 


23.4 


59.44 


6686 


11.493 


1655.0 


23.6 


59.95 


6454 


11.592 


1669.3 


23.8 


60.45 


6225 


11.690. 


1683.3 


24.0 


60.96 


5997 


11.788 


1697.4 


24.2 


61.47 


5771 


11.886 


1711.6 


24.4 


61.98 


5547 


11.984 


1725.7 


24.6 


62.48 


5325 


12.083 


1739.9 


24.8 


62.99 


5105 


12.182 


1754.0 


25.0 


63.50 


4886 


12.280 


1768.2 


25.2 


64.01 


4670 


12.377 


1782.3 


25.4 


64.52 


4455 


12.475 


1796.5 


25 6 


65.02 


4241 


12.573 


1810.7 


1 25.8 


65.53 


4030 


12.671 


1824.8 



66 



ENGINEERING THERMODYNAMICS 



TABLE IX Continued 



Standard Barometer. 


Altitude, Feet above 
Sea Level. 


Pressure, Pounds per 


Inches. 


Centimeters. 


Square Inch. 


Square Foot. 


26.0 


65.04 


3820 


12.770 


1838.9 


26.1 


66.30 


3715 


12.819 


1846.0 


26.2 


66.55 


3611 


12.868 


1853.1 


26.3 


66.80 


3508 


12.918 


1860.2 


26.4 


67.06 


3404 


12.967 


1867.3 


26.5 


67.31 


3301 


13.016 


1874.3 


26.6 


67.57 


3199 


13.065 


1881.4 


26.7 


67.82 


3097 


13.113 


1888.5 


26.8 


68.08 


.2995 


13 163 


1895.5 


26.9 


68.33 


2894 


13.212 


1902.6 


27.0 


68.58 


2793 


13.261 


1909.7 


27.1 


68.84 


2692 


13.310 


1916.7 


27.2 


69.09 


2592 


13.359 


1923.8 


27.3 


69.34 


2493 


13.408 


1930.9 


27.4 


69.60 


2393 


13.457 


1938.0 


27.5 


69.85 


2294 


13.507 


1945.1 


27.6 


70.10 


2195 


13.556 


1952.1 


27.7 


70.35 


2097 


13 . 605 


1959.2 


27.8 


70.61 


1999 


13.654 


1966.3 


27.9 


70.87 


1901 


13.704 


1973 . 3 


28.0 


71.12 


1804 


13.753 


1980.4 


28.1 


71.38 


1707 


13.802 


1987.5 


28.2 


71.63 


1610 


13.850 


1994.5 


28.3 


71.88 


1514 


13.899 


2001 . 6 


28.4 


72.14 


1418 


13.948 


2008.7 


28.5 


72.39 


1322 


13.998 


2015.7 


28.6 


72.64 


1227 


14.047 


2022.8 


28.7 


72.90 


1132 


14.096 


2030.0 


28.8 


73.15 


1038 


14.145 


2037.0 


'28.9 


73.40 


943 


14.194 


2044.1 


29.0 


73.66 


849 


14.243 


2051.2 


29.1 


73.92 


756 


14.293 


2058.2 


29.2 


74.16 


663 


14.342 


2065.3 


29.3 


74.42 


570 


14.392 


2072 . 4 


29.4 


74.68 


477 


14.441 


2079 . 4 


29.5 


74.94 


384 


14.490 


2086.5 


29.6 


75.18 


292 


14.539 


2093.6 


29.7 


75.44 


261 


14.588 


2100.7 


29.8 


75.69 


109 


14.637 


2107.7 


29.9 


75.95 


+18 


14.686 


2114.7 


29.92 


76.00 





14.696 


2116.1 


30.0 


76.20 


- 73 


14.734 


2121.7 


30.1 


76.46 


-163 


14.783 


2128.8 


30.2 


76.71 


-253 


14.833 


2135.9 


30.3 


76.96 


-343 


14.882 


2143.0 


30.4 


77.22 


-433 


14.931 


2150.1 


30.5 


77.47 


-522 


14.980 


2157.2 


30.6 


77.72 


-611 


15.030 


2164.2 


30.7 


77.98 


-700 


15.078 


2171.3 


30.8 


78.23 


-788 


15.127 


2178.4 


30.9 


78.48 


-877 


15.176 


2185.5 


31.0 


78.74 


-965 


15.226 


2192.6 



TABLES 



67 



TABLE X 

VALUES OF s IN THE EQUATION PV S = CONSTANT FOR VARIOUS SUBSTANCES 

AND CONDITIONS 



Substance. 




s 


Remarks or Authority. 


All gases 


Isothermal 


1 1 




All gases and vapors. . 
All saturated vapors . . 
All gases and vapors . . 
Air 


Constant pressure 
Isothermal 
Constant volume 
Adiaba'ic 





00 

1.4066 


Accepted thermody- 
narnic law 

Smithsonian Tables 


Air 


Compressed in cylinder 


1 4 


Experience 


iAjnmonia (NH 3 ) 
Ammonia (NH 3 ) 
Bromine . . . 


Adiabatic, wet 
Adiabatic, superheated 
Adiabatic 


1.1 
1.3 
1.293 


Average 
Thermodynamics 
Strecker 


Carbon dioxide (CO 2 ) . 
Carbon monoxide (CO) 
Carbon disulphide 
(CSz) 


Adiabatic 
Adiabatic 

Adiabatic 


1.300 
1.403 

1.200 


Rontgen, Wullner 
Cazin, Wullner 

Beyne 


Chlorine (Cl) 


Adiabatic 


1.323 


Strecker 


Chloroform 
(CC1 3 CH(OH) 2 ).... 
Ether (C 2 H 5 OC 2 H 5 ). . . 
Hydrogen (H 2 ) 


Adiabatic 
Adiabatic 
Adiabatic 


1.106 
1.029 
1 410 


Beyne, Wullner 
Muller 
Cazin 


Hydrogen sulph . (H 2 S) 
Methane (CH 4 ) 
Nitrogen (N 2 ) 
Nitrous oxide (NO 2 ) . . 
Pintsch gas 


Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 
Adiabatic 


1.276 
1.316 
1.410 
1.291 
1'.24 


Muller 
Muller 
Cazin 
Wullner 
Pintsch Co. 


Sulphide diox (SO 2 ) ... 
Steam, superheated . .. 
Steam, wet 

Steam wet 


Adiabatic 
Adiabatic 
Adiabatic 

Adiabatic 


1.26 
1.300 
Variable 

1.111 


Cazin, Muller 
Smithsonian Tables 
(From less than 1 to 
more than 1.2) 
Rankine , 


Steam, wet 


Adiabatic 


1+14X% moist. 


Perry 


Steam, wet 


Adiabatic 


1.035 + .1X% moist. 


Gray 


Steam, wet 
Steam, dry 


Expanding in cylinder 
Saturation law 


1. 
1.0646 


Average from practice 
Begnault 











ENGINEERING THERMODYNAMICS 

TABLE XI 
HORSE-POWER PER POUND MEAN EFFECTIVE PRESSURE. 

aS 



VALUE OF K e = 



33000 



Diameter 
of 
Cylinder, 
Inches. 


Speed of Piston in Feet per Minute. 


100 


200 


300 


400 


500 


600 


700 


800 


900 


4 


0.0381 


0.0762 


0.1142 


0.1523 


0.1904 


0.2285 


0.2666 


0.3046 


0.3427 


4* 


0.0482 


0.0964 


0.1446 


0.1928 


0.2410 


0.2892 


0.3374 


0.3856 


0.4338 


5 


0.0592 


0.1190 


0.1785 


0.2380 


0.2975 


0.3570 


0.4165 


0.4760 


0.5355 


5* 


0.0720 


0.1440 


0.2160 


0.2880 


0.3600 


0.4320 


0.5040 


0.5760 


0.6480 


6 


0.0857 


0.1714 


0.2570 


0.3427 


0.4284 


0.5141 


0.5998 


0.6854 


0.7711 


61 


0.1006 


0.2011 


0.3017 


0.4022 


0.5028 


0.6033 


0.7039 


0.8044 


0.9050 


7 


0.1166 


0.2332 


0.3499 


0.4665 


0.5831 


0.6997 


0.8163 


0.9330 


1.0490 


n 


0.1339 


0.2678 


0.4016 


0.5355 


0.6694 


0.8033 


0.9371 


1.0710 


1.2049 


8 


0.1523 


0.3046 


0.4570 


0.6093 


0.7616 


0.9139 


1.0662 


1.2186 


1.3709 


8| 


0.1720 


0.2439 


0.5159 


0.6878 


0.8598 


1.0317 


1.2037 


1.3756 


1.5476 


9 


0.1928 


0.3856 


0.5783 


0.7711 


0.9639 


1 . 1567 


1.3495 


1.5422 


1 . 7350 


9* 


0.2148 


0.4296 


0.6444 


0.8592 


1.0740 


1.2888 


1.5036 


1.7184 


1.9532 


10 


0.2380 


0.4760 


0.7140 


0.9520 


1 . 1900 


1.4280 


1.6660 


1.9040 


2.1420 


11 


0.2880 


0.5760 


0.8639 


1.1519 


1.4399 


1 . 7279 


2.0159 


2 . 3038 


2.5818 


12 


0.3427 


0.6854 


1.0282 


1.3709 


1.7136 


2.0563 


2.3990 


2.7418 


3.0845 


13 


0.4022 


0.8044 


1.2067 


1.6089 


2.0111 


2.4133 


2.8155 


3.2178 


3.6200 


14 


0.4665 


0.9330 


1.3994 


1 . 8659 


2.3324 


2.7989 


3.2654 


3.7318 


4.1983 


15 


0.5355 


1.0710 


1.6065 


2.1420 


2.6775 


3.2130 


3.7485 


4.2840 


4.8195 


16 


0.6093 


1.2186 


1.8278 


2.4371 


3.0464 


3.6557 


4.2650 


4.8742 


5.4835 


17 


0.6878 


1.2756 


1.9635 


2.6513 


3.3391 


4.0269 


4.6147 


5.4026 


6.1904 


18 


0.7711 


1.5422 


2.3134 


3.0845 


3.8556 


4.6267 


5.3987 


6.1690 


6.4901 


19 


0.8592 


1.7184 


2.5775 


3.4367 


4.2858 


5.1551 


6.0143 


6.8734 


7.7326 


20 


0.9520 


1.9040 


2.8560 


3.8080 


4.7600 


5.7120 


6.6640 


7.6160 


8.5680 


21 


1.0496 


2.0992 


3.1488 


4.1983 


5.2475 


6.2975 


7.3471 


8.3966 


9.4462 


22 


.1519 


2.3038 


3.4558 


4.6077 


5.7596 


6.9115 


8.0643 


9.2154 


10.367 


23 


.2590 


2.5180 


3.7771 


5.0361 


6.2951 


7.5541 


8.8131 


10.072 


11.331 


24 


.3709 


2.7418 


4.1126 


5.4835 


6.8544 


8.2253 


9.5962 


10.967 


12.338 


25 


.4875 


2.9750 


4.4625 


5.9500 


7.4375 


8.9250 


10.413 


11.900 


13.388 


26 


.6089 


3.2178 


4.8266 


6.4355 


8.0444 


9.6534 


11.262 


12.871 


14.480 


27 


.7350 


3.4700 


5.2051 


6.9401 


8.6751 


10.410 


12.145 


13.880 


15.615 


28 


.8659 


3.7318 


5.5978 


7.4637 


9.3296 


11.196 


13.061 


14.927 


16.793 [ 


29 


2.0016 


4.0032 


6.0047 


8.0063 


10.008 


12.009 


14.011 


16.013 


18.014 


30 


2.1420 


4.2840 


6.4260 


8.5680 


10.710 


12.852 


14.994 


17.136 


19.278 [ 


31 


2.2872 


4.5744 


6.8615 


9.1487 


11.436 


13.723 


16.010 


18.287 


20.585 [ 


32 


2.4371 


4.8742 


7.3114 


9.7485 


12.186 


14.623 


17.060 


19.497 


21.934 1 


33 


2.5918 


5.1836 


7.7755 


10.367 


12.959 


15.551 


18.143 


20.735 


23.326 I 


34 


2.7513 


5.5026 


8.2538 


11.005 


13.756 


16.508 


19.259 


22.010 


24.762 I 


35 


2.9155 


5.8310 


8.7465 


11.662 


14.578 


17.493 


20.409 


23.224 


26.240 | 


36 


3.0845 


6.1690 


9.2534 


12.338 


15.422 


18.507 


21.591 


24.676 


27.760 1 


37 


3.2582 


6.5164 


9.7747 


13.033 


16.291 


19.549 


22.808 


26.066 


29.324 1 


38 


3.4367 


6.8734 


10.310 


13 . 747 


17.184 


20.620 


24.057 


27.494 


30.930 1 


39 


3.6200 


7.2400 


10.860 


14.480 


18.100 


21.720 


25.340 


28.960 


32.580 



TABLES 
TABLE XI Continued 



69 



Diameter 
of 


Speed of Piston in Feet per Minute. 


Cylinder, 
Inches. 


100 


200 


300 


400 


500 


600 


700 


800 


900 


40 


3.8080 


7.6160 


11.424 


15.232 


19.040 


22.848 


26.656 


30.464 


34.272 


41 


4.0008 


8.0016 


12.002 


16.003 


20.004 


24.005 


28.005 


32.006 


36.007 


42 


4.1983 


8.3866 


12.585 


16.783 


20.982 


25.180 


29.378 


33.577 


37.775 


43 


4.4006 


8.8012 


13.202 


17.602 


22.003 


26.404 


30.804 


35.205 


39.606 


44 


4.6077 


9.2154 


13.823 


18.431 


23.038 


27.646 


32.254 


36.861 


41.469 


45 


4.8195 


9.6390 


14.459 


19.278 


24.098 


28.917 


33.737 


38.556 


43.376 


46 


5.0361 


10.072 


15.108 


20.144 


25.180 


30.216 


35.253 


40.289 


45.325 


47 


5.2574 


10.515 


15.772 


21.030 


26.287 


31.545 


36.802 


42.059 


47.317 


48 


5.3845 


10.967 


16.451 


21.934 


27.418 


32.901 


38 . 385 


43.868 


49.352 


49 


5.7144 


11.429 


17.143 


22.858 


28.572 


34.286 


40.001 


45.715 


51.429 


50 


5.9500 11.900 


17.850 


23.800 


29.750 


35.700 


41.650 


47.600 


53.550 


51 


6. 1904 | 12.381 


18.571 


24.762 


30.952 


37.142 


43.333 


49.523 


55.713 


52 


6.4355 


12.871 


19.307 


25.742 


32.178 


38.613 


45.049 


51.484 


57.920 


53 


6.6854 


13.371 


20.056 


26.742 


33.427 


40.113 


46 . 798 


53.483 


60.169 


54 


6.9401 


13 . 880 


20.820 


27.760 


34.700 


41.640 


48.581 


55.521 


62.461 


55 


7.1995 


14.399 


21.599 


28.798 


35.998 


43.197 


50.397 


57.596 


64.796 


56 


7.4637 


14.927 


22.391 


29.855 


37.318 


44.782 


52.246 


59.709 


67.173 


57 


7.7326 


15.465 


23.198 


30.930 


38.663 


46.396 


54.128 


61.861 


69.594 


58 


8.0063 


16.013 


24.019 


32.025 


40.032 


48.038 


56.044 


64.051 


72.057 


59 


8.2849 


16.570 


24.854 


33.139 


41.424 


49.709 


57.993 


66.278 


74.563 


60 


8.5680 


17.136' 25.704 


34.272 


42.840 


51.408 


59.976 


68.544 


77.112 




- 1 















70 



ENGINEERING THERMODYNAMICS 



GENERAL FORMULA RELATING TO PRESSURE- VOLUME 
CALCULATIONS .OF WORK AND POWER 

Work = W = Force X Distance =FxL; 

= Pressure xArea X Distance =P XA XZ>; 
= Pressure X Volume change =Px(V 2 V l ). 
Force of acceleration =mass X acceleration. 



-. 

g di 
Work of acceleration = mass X difference of (velocity) 2 , 



1 w 

__ 



w 

__ 



,,. 

_,*). 



Velocity due to work of acceleration = square root of the sum of (initial velocity) 
plus 20Xwork per pound of substance accelerated. 



w 



or if initial velocity is zero, 



w 



w 



Pressure Volume Relation for Expansion or Compression 
PV S ^PiVi 8 =P 2 V 2 S =K, a constant. (See Table VIII for values of s.) 



log ~ 



(Note graphical method for finding s when variable, see text.) 



FORMULA RELATING TO PRESSURE- VOLUME CALCULATIONS 71 



Work done during a pressure volume change represented by the equation (PV 8 K) 
tween points represented by 1 and 2 in figure =area under curve, Wi. 



If 8=1, 



rvi C V2 dV 

= I PdV=Kl 
JVi J Vi V s 



= 1. 



If s is not equal to 1, 



Oi 



p.r.r, /PA 
^iL^W Jj 



Work of admission, complete expansion and exhaust for engines = area to left of 
curve (see figure) TF 2 . Same for admission, compression and expulsion for com- 
pressors. Both cases without clearance. 

(When s = l, Wz is same as area under curve, W\ (see above). 






Clearance, expressed as a fraction of displacement =c, as a volume 



72 



ENGINEERING THERMODYNAMICS 



Cl D D \P 



Indicated horse-power = (mean effective pressure, pounds per square inch X effect! 1 
area of piston, square itches X length of stroke, feetXnumber of working cycles pei 
formed per minute) divided by 33,000. 



T TIP ='' 

33000 
Specific displacement = displacement in one direction for one side of a piston, u 

cu.ft. per hour per H.P. =D S = 13,750, z 

(m.e.p.) 

where z is the number of strokes required o compl e one cycle. 

Velocity of a jet due to its own expansion =the square root of the product of 2g Xworl 
done by admission, complete expansion and exhaust of 1 Ib. of the substance. 



Weight of flow through nozzle or orifices, pounds per second =w = (velocity, feet 
second X area, square feet, of orifice) -f- (volume per pound of substance at section where 
area is measured). 



1 ( s-l ) 

W= ^A = SQ2 ( ' }' \-P V fl-^ ^~*~1 f 



Maximum discharge w for a given initial pressure occurs when 






CHAPTER II 



VORK OF COMPRESSORS, HORSE-POWER AND CAPACITY OF AIR, GAS AND 
VAPOR COMPRESSORS, BLOWING ENGINES AND DRY VACUUM PUMPS. 

1. General Description of Structure and Processes. There is quite a 
arge class of machines designed to receive a cylinder full of some gas at one 
Constant pressure and after the doing of work on the gas through decreasing 
Volumes and rising pressures, to discharge the lesser volume of gas against a 
Constant higher pressure. These machines are in practice grouped into sub- 
classes, each having some specific distinguishing characteristic. For example, 
plowing engines take in air at atmospheric pressure or as nearly so as the 
ralve and port resistance will permit, and after compression deliver the air 
it a pressure of about three atmospheres absolute for use in blast furnaces. 
These blowing engines are usually very large, work at low but variable speeds, 
-mt always deliver against comparatively low pressures; they, therefore, have 
he characteristics of large but variable capacity and low pressures. A great 
^ariety of valves and driving gears are used, generally mechanically moved 
uction and automatic spring closed discharge valves, but all valves may be 
oitomatic. The compressor cylinder is often termed the blowing tub and the 
ompressed or blast air frequently is spoken of as wind by furnace men. They 
ire all direct-connected machines, an engine forming with the compressor one 
nachine. The engine formerly was always of the steam type, but now a change 
b being affected to permit the direct internal combustion of the blast furnace 
vaste gases in the cylinders of gas engines. These gas-driven blowing engines, 
'ihowing approximately twice the economy of steam-driven machines, will in 
:ime probably entirely displace steam in steel plants, and this change will take 
>lace in proportion to the successful reduction of cost of repairs, increase of 
t : eliability and life of the gas-driven blowing engines to equal the steam-driven, 
feme low-pressure blowers are built on the rotary plan without reciprocating 
bistons, some form of rotating piston being substituted, and these, by reason 
f greater leakage possibilities, are adapted only to such low pressures as 5 Ibs. 
>er square inch above atmosphere or thereabouts. These blowers are coming 
tito favor for blasting gas producers, in which air is forced through thick coal 
>eds either by driving the air or by drawing on the gas produced beyond the 
>ed. They are also used for forcing illuminating gas in cities through pipes 
>therwise too small, especially when the distances are long. In general very 
ow pressures and large capacities are the characteristics of the service whether 
he work be that of blowing or exhausting or both. For still lower pressures, 
neasurable by water or mercury columns, fans are used of the disk or propeller 

73 



74 ENGINEERING THERMODYNAMICS 

or centrifugal type. These fans are most used for ventilation of buildings an 
mines, but a modification, based on the principles of the steam turbine reversec 
and termed turbo-compressors, is being rapidly adapted to such higher pres 
sures as have heretofore required piston compressors. 

When high-pressure air is required for driving rock drills in mines and fc 
hoisting engines, for tools, as metal drills, riveters, chipping chisels, for ca 
air brakes, the compressors used to provide the air are termed simply ai 
compressors. These compressors usually take in atmospheric air and compres 
it to the desired pressure, the capacity required being usually adjustable 
they have valves of the automatic type throughout commonly, but in larg 
sizes frequently are fitted with mechanically operated suction valves t 
decrease the resistance to entrance of air and so increase economy, a com 
plication not warranted in small machines. When the pressures of deliver 
are quite high the compression is done in stages in successive cylinders, th 
discharge from the first or low-pressure cylinder being delivered through 
water cooler or inter cooler to the second cylinder and occasionally to a thir 
in turn. This staging with intermediate or intercooling results in bettc 
economy, as will be seen later in detail, and permits the attainment of th 
desired quantity of cool compressed air for subsequent use with the expenditur 
of less work, the extra complication and cost being warranted only whe 
machines are large and final pressures high. 

In the operation of large steam condensers, non-condensible gases wi 
collect and spoil the vacuum, which can be maintained only by the continuoi 
removal of these gases, consisting of air, carbon dioxide and gases of anim; 
and vegetable decomposition originally present in the water. When these gas< 
are separately removed the machine used is a special form of compressor terme 
a dry vacuum pump which, therefore, receives a charge at the absolute pre 
sure corresponding to the vacuum, or as nearly so as the entrance resistano 
permits, and after compression discharges into the atmosphere at a pressui 
in the cylinder above atmosphere equivalent to discharge resistance. Natura 
gas wells near exhaustion can sometimes be made to flow freely by the applic. 
tion of a compressor capable of drawing a charge at a pressure below atmospher; 
but whether the charge be received below atmospheric pressure or above i 
in normal wells, the compressor will permit the delivery of the gas to distal 
cities or points of consumption even 250 miles away through smaller pip 
than would be otherwise possible. Natural-gas compressors, some steam- ar 
some gas-engine driven, are in use for both these purposes, compressing natur 
gas from whatever pressure may exist at the well to whatever is desired at tl 
beginning of the pipe line. 

In the preparation of liquid ammonia or carbonic acid gas for the marke 
as such, or in the operation of refrigerating machinery, wet or dry vapor is con 
pressed into a condenser to permit liquefaction by the combined effect of hij 
pressure and cooling. One form of refrigerating machine merely compress 
air, subsequently expanding it after preliminary cooling by water, so th; 
after expansion is complete it will become extremely cold. 




WOEK OF COMPRESSORS 75 

All these compressing machines have, as a primary purpose, either the 

>moval of a quantity of low-pressure gases from a given place, or the delivery 

f a quantity of higher-pressure gas to another place or both, but all include 

Dmpression as an intermediate step between constant-pressure admission and 

onst ant-pressure discharge as nearly as structure may permit. They will all 

ivolve the same sort of physical operations and can be analyzed by the same 

rinciples except the wet-vapor or wet-gas compressors, in which condensation 

|<r evaporation may complicate the process and introduce elements that can 

Jje treated only by thermal analysis later. Safe compressors cannot be built 

rdth zero cylinder clearance, hence at the end of delivery there will remain in 

ifhe clearance space a volume of high-pressure gases equal to the volume of the 

(Clearance space. On the return stroke this clearance volume will expand until 

|he pressure is low enough to permit suction, so that the new charge cannot 

nter the cylinder until some portion of the stroke has been covered to permit 

''his re-expansion of clearance gases. 

It is quite impossible to study here all the effects or influences of structure 
Ita indicated by the compressor indicator cards, but a quite satisfactory treat- 
nent can be given by the establishment of reference diagrams as standards of 
Comparison and noting the nature of the differences between the actual cases and 
'he standard reference diagram. These standard reference diagrams will really 
>e pressure-volume diagrams, the phases of which correspond to certain hypoth- 
*!ses capable of mathematical expression, such as constant pressure, constant 
Volume, expansion, and compression, according to some law, or with some 
definite value of s fixing either the heat-exchange character of the process or 
fihe substance, as already explained. 

2. Standard Reference Diagrams or PV Cycles for Compressors and Methods 
)f Analysis of Compressor Work and Capacity. All the standard reference 
diagrams will include constant-pressure lines corresponding to delivery and 
'supply at pressures assumed equal to whatever exists outside the. v cylinder on 
lather delivery or suction side, that is, assuming no loss of pressure on delivery 
br suction. The compression may be single or multi-stage with various 
jimounts of cooling in the intercooler, but in multi-stage compression 
ohe standard reference diagram will be assumed to involve intermediate 
jooling of the gases to their original temperature, so that the gases 
3ntering all cylinders will be assumed to have the same temperature and 
X) maintain it constant during admission. Another difference entering into 
Ijhe classification of standard reference diagrams is the laws of compression 
is defined by the exponent s. Integration of the differential work expres- 
sion will take a logarithmic form for s = l, and an exponential form for all 
Uther values, thus giving two possible reference compression curves and two 
sets of work equations. 

(a) The isothermal f or which s = 1 , no matter what the gas, and which is 
the consequence of assuming that all the heat liberated by compression is con- 
tinuously carried away as fast as set free, so that the temperature cannot rise 
it all. 



76 ENGINEERING THERMODYNAMICS 

(6) The exponential for which s has a value greater than one, general! 
different for every gas, vapor or gas-vapor mixture, but constant for any on 
gas, and also for dry vapors that remain dry for the whole process. Wet vapor 
having variable values of s cannot be treated by the simple pressure-volum 
analysis that suffices for the gases, but must be analyzed thermally. Th 
adiabatic value of s is a consequence of assuming no heat exchange at a 
between the gas and anything else and is a special case of the general exponer 
tial class. 

Just why these two assumptions of thermal condition should result in th 
specified values of s will be taken up under the thermal analysis part of thi 
work, and is of no interest at this time. 

As a consequence of these phase possibilities there may be established eigh 
standard reference diagrams or pressure-volume cycles defined by their phase* 
as shown in Fig. 23, four for single-stage compression and two each for two an 
three stages. These might be extended by adding two more for four stage 
and so on, but as it seldom is desirable, all things being considered, to go beyon 
three, the analysis will stop with the eight cycles or reference diagrams showr 

SINGLE-STAGE COMPRESSION REFERENCE CYCLES OR PV DIAGRAMS 

Cycle 1. Single-stage Isothermal Compression without Clearance. 
Phase (a) Constant pressure supply. 
" (6) Isothermal compression. 
'" (c) Constant pressure delivery. 
" (d) Constant zero-volume pressure drop. 
Cycle 2. Single-stage Isothermal Compression with Clearance. 
Phase (a) Constant pressure supply. 
" (6) Isothermal compression. 
" (c) Constant pressure delivery. 
" (d) Isothermal re-expansion. 

Cycle 3. Single-stage Exponential Compression without Clearance. 
Phase (a) Constant pressure supply. 
" (6) Exponential compression. 

(c) Constant pressure delivery. 
" (d) Constant zero-volume pressure drop. 
Cycle 4. Single-stage Exponential Compression with Clearance. 
Phase (a) Constant pressure supply. 
(b) Exponential compression. 
" (c) Constant pressure delivery. 
" (d) Exponential re-expansion. 



WORK OF COMPRESSORS 



77 





123456 
Volumesja Cubic Feet 








56 123456 

aic Feet Volumes ia Cubic Feet 

rs; One-, Two-, and Three-Stage, with and without Clearance, 
onential Laws. 


































y 


















7 


















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r 










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l\ 




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345 
Cubic Feet 

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s 

spunod 


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glumes in Cubic Feet 








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78 ENGINEERING THERMODYNAMICS 



MULTI-STAGE COMPRESSION 

The phases making up multi-stage compression cycles may be consider 
two ways, first, as referred to each cylinder and intercooler separately, b 
second, as referred to the pressure volume changes of the gases themselve 
regardless of whether the changes take place in cylinders or intercoolers. 

For example, if 10 cu.-ft. of hot compressed air be delivered from the firs 
cylinder of 50 cu.-ft. displacement, the phase referred to this cylinder is a con 
stant-pressure decreasing volume, delivery line whose length is ^ of the whoL 
diagram, exactly as in single-stage compression. If this 10 cu.-ft. of air delivj 
ered to an intercooler became 8 ft. at the same constant pressure as the firs 
cylinder delivery, the phase would be indicated by a constant-pressure volum 
reduction line 2 cu.-ft long to scale, or referred to the original volume of ai 
admitted to the first cylinder, a line gV of its length. Finally, admitting thi 
8 cu.-ft of cool air to the second cylinder and compressing it to | of its vo 
ume would result in a final delivery line at constant pressure of a length of i cj 
the length of the second cylinder diagram, but as this represents only 8 cu.-ft| 
the final delivery will represent only |X8 = 1.6 cu.-ft. This 1.6 cu.-ft. wilj 
when referred to the original 50 cu.-ft. admitted to the first cylinder, be repr< 

1,6 

sented by a constant-pressure line, "rr = .032, of the whole diagram lengtl 

oU 

which in volume is equivalent to -J- of the length of the second cylind<: 
diagram. It should be noted also that three volume change operations tali 
place at the intermediate pressure; first, first cylinder delivery; secon< 
volume decrease due to intercooling; third, second cylinder admission, tl 
net effect of which referred to actual gas volumes, regardless of place whe 
the changes happen, is represented by the volume decrease due to inte 
cooling only; A diagram of volumes and pressures representing the resultai 
of all the gas processes is called in practice the combined PV diagram for tl 
two cylinders, or when plotted from actual indicator cards with due regai 
for the different clearances of each cylinder the combined indicator diagran 
It is proper in the study of the whole process of compression to consider tl 
cycle consisting of phases referred to true gas volumes rather than phas, 
referring to separate cylinder processes, which is equivalent to imagining tlj 
whole cycle carried out in one cylinder. 

Intercooling effects measured by the amount of decrease of volume j 
constant pressure will, of course, depend on the amount of cooling or redu 
tion of temperature, but in establishing a standard reference diagram sorj 
definite amount capable of algebraic description must be assumed as . 
intercooling hypothesis. 

It has already been shown, Fig. 6, Chapter I, that from any original stf * 
of pressure and volume the exponential and isothermal could be drawn, divergi fc 
an amount depending on the difference between the defining exponent, L 
If, after reaching a given state on the exponential curve, the gas be cooled |J 



WORK OF COMPRESSORS 79 

mstant pressure to its original temperature, the point indicating its condition 
ill lie by definition on the other curve or isothermal and the cooling process 
represented by a horizontal joining the two curves. Such intercooling as 
lis will be defined as perfect intercooling, for want of a better name, and its 
ressure-volume effects can be treated by the curve intersections. It is now 
ossible to set down the phases for the standard reference diagrams of multi- 
age compression, if in addition to the above it be admitted, as will be proved 
,ter, that there is a best or most economical receiver pressure. 



TWO-STAGE COMPRESSION REFERENCE CYCLES OR PV DIAGRAMS 

Cycle 5. Two-stage Exponential Compression without Clearance, Perfect 

Intercooling at Best Receiver Pressure. 
Phase (a) Constant pressure supply. 

' ' (b) Exponential compression to best receiver pressure. 
' ' (c) Constant pressure perfect intercooling of delivered gas. 
' ' (d) Exponential compression from best receiver pressure. 
' ' (e) Constant pressure delivery. 
' ' (/) Constant zero-volume pressure drop. 
Cycle 6. Two-stage Exponential Compression with Clearance, Perfect 

Intercooling at Best Receiver Pressure. 
Phase (a) Constant pressure supply. 

' ' (6) Exponential compression to best receiver pressure. 
1 ' (c) Constant pressure perfect intercooling of delivered gas. 
' ' (d) Exponential re-expansion of first stage clearance. 
' ' (e) Exponential compression from best receiver pressure. 
' ' (/) Constant pressure delivery. 
' ' (g) Exponential re-expansion of second stage clearance. 



THREE-STAGE COMPRESSION REFERENCE CYCLES OR PV DIAGRAMS 

Cycle 7. Three-stage Exponential Compression, without Clearance, Per- 
fect Intercooling at Best Two Receiver Pressures. 
Phase (a) Constant pressure supply. 
1 ' (b) Exponential compression to first receiver pressure. 

(c) Perfect intercooling at best first receiver presssure, 

(d) Exponential compression from best first to best second 

receiver pressure. 

' ' (e) Perfect intercooling at best second receiver pressure. 
' ' (/) Exponential compression from best second receiver pressure. 
' ' (g) Constant pressure delivery. 
" (h) Constant zero-volume pressure drop. 



80 ENGINEERING THERMODYNAMICS 

Cycle 8. Three-stage Adiabatic Compression with Clearance, Perfec 

Inter cooling at Best Two Receiver Pressures. 
Phase (a) Constant pressure admission. 

' ' (6) Exponential compression to best first receiver pressure. 
' ' (c) Perfect cooling of delivered gas at best first receiver pressi 
' ' (d) Exponential re-expansion of first stage clearance. 
" (e) Exponential compression from best first to best sec< 

receiver pressure. 
1 ' (/) Perfect intercooling of delivered gas at best second recer 

pressure. 

' l (g) Exponential re-expansion of second stage clearance. 
' ' (h) Exponential compression from best second receiver pressi 
' l (i) Constant pressure delivery. 
' ' (f) Exponential re-expansion of third stage clearance. 

It should be noted that cycles 6 and 8 may be sub-divided into any numbe 
of cases, of which some of the most characteristic are shown: (a) where 
.clearance volume in each cylinder bears the same ratio to the displacerru 
of that cylinder, and commonly called equal clearances; (6) where the clearai 
are. such that the volume after re-expansion in the higher-pressure cylin< 
is equal to the volume of clearance in the next lower-pressure cylinder, caus 
the combined diagram to have a continuous re-expansion line, a case wl 
may be called proportionate clearance; and (c) the general case in which tl 
is no particular relation between clearances in the several cylinders. 

By means of these definitions or their mathematical equivalents in syml 
it will be possible to calculate work as a function of pressures and volui 
and by various transformations of a general expresssion for work of a referei 
cycle to calculate the horse-power corresponding to the removal of a gr 
volume of gas per minute from the low-pressure supply or to the delivery 
another volume per minute to the high-pressure receiver or per unit weij 
per minute. It will also be possible to calculate the necessary cylinder si; 
or displacement per unit of gas handled, and the horse-power necessary 1| 
drive the compressing piston at a specified rate and further to calculate tl! 
work and horse-power of cylinders of given size and speed. In order that thei 
calculations of a numerical sort may be quickly made, which is quite necessaij 
if they are to be useful, the formulas must be definite and of proper form, tl 
form being considered proper when little or no algebraic transformation 
necessary before numerical work is possible. While special expressions f<! 
each case are necessary to facilitate numerical work, it is equally important 
if not more so, to make clear the broad general principles or methods of attac. 
because it is quite impossible to set down every case or even to conceive at tlj 
time of writing of all different cases that must in future arise. The treaj 
ment, then, must be a combination of general and special, the general metho- I 
being applied successively, to make them clear and as a matter of drill, nj 
to every possible case, but only to certain characteristic or type forri 



WORK OF COMPRESSORS 81 

,f cases, such as are here set down as standard reference diagrams. 
ndividual cases may be judged by comparison with these and certain factors 
f relation established which, being ratios, may be and are called efficiencies. 
.1ms, if a single-stage compressor should require two horse-power per cubic 
oot of tree air compressed per minute, and Cycle I should for the same 
>ressure limits require only one horse-power for its execution, then the efficiency 
>f the real compression would be 50 per cent referred to Cycle I, and similar 
actors or efficiencies for other compressors similarly obtained; a com- 
parison of the factors will yield information for a judgment of the two 
:ompressors. 

In what follows on the work and gas capacity of compressors two methods 
j)f attack will be used. 

1. General pressure- volume analysis in terms of gas pressures and volumes 
esulting in the evaluation of work per cubic foot of low- or high- pressure 

i;aseous substance. 

2. Transformation of results of (1) to yield volumetric efficiencies, mean 
.effective pressures, work, horse-power, and capacity in terms of dimensions 
of cylinders and clearances. 

3. Single-stage Compressor, No Clearance, Isothermal Compression 
[Cycle 1), Work, Capacity, and Work per Cubic Foot in Terms of Pressures 
ind Volumes. The standard reference diagram is represented by Fig. 24, on 
tfhich the process (A to B) represents admission or supply at constant pressure; 
'B to C) compression at constant temperature; (C to D) delivery at constant 
pressure; and (D to A) 2ero-volume. 

Let F& = The number of cubic feet of low pressure gas in the cylinder after 

admission, represented to scale on the diagram by AB and equal 

to the volume at B\ 
V c = volume in cubic feet of the gas in cylinder when discharge begins, 

represented by DC, which is the volume at C; 
" Pb = absolute pressure in pounds per square foot, at which supply 

enters cylinder = (Sup. Pr.) = pressure at B; 
pb = Pb+ 144 = absolute supply pressure in pounds per square inch = 

(sup.pr.); 
PC absolute pressure in pounds per square foot, at which delivery 

occurs = (Del. Pr.) = pressure at C; 
PC = PC -5- 144 = absolute delivery pressure in pounds per square inch 

= (del.pr.); 
p 

R p = = ratio of delivery pressure to supply pressure; v, 

^ 



W = foot-pounds work done for the cycle; \ 

(H. P. Cap.) = volume of gas delivered in cubic feet per cycle, at 

temperature same as that of supply; 
(L. P. Cap. )= volume of gas drawn into cylinder, cubic feet per cycle. 

For this no clearance case (L. P. Cap.) = F&. 




82 



ENGINEERING THERMODYNAMICS 



Referring to Fig. 24, the work for the cycle is the sum of compression anc 
delivery work, less admission work, or by areas 



Net work A BCD = compression work EBCG+ deli very work GCDF 

admission work EBAF 
Algebraically this is equivalent to 




1 A 

FIG. 24. One-stage Compressor Cycle 1, No Clearance, Isothermal. 



But since P c V c = PbVb the expression becomes 



which is the work for the execution of the cycle when pressures and volui 
are in pounds per square foot, and cubic feet. The equivalent expressh 
for pounds per square inch and cubic feet is 



W =144 



Pb 



WORK OF COMPRESSORS 83 

Since, when there is no clearance the volume taken into the cylinder for 
ach cycle (L. P. Cap.) is equal to the volume at B, Vb, the expression Eq. (29) 
aay be stated thus, symbolic form. 

TF=144(sup.pr.)(L. P. Cap.) loge R P (30) 

The work per cubic foot of low pressure gas, foot-pounds, will be the above 
xpression divided by (L. P. Cap.), or 

I W 

144 (sup.pr.) log, R p (31) 



(L. P. Cap.) 
The work per cubic foot of high-pressure gas delivered will be 

(H P ^~~ ^ = "^ ( su P*P r ') Rp ^& e RP> 
mce 
>r 



vhich expressed symbolically is 

.) = (H.P.Cap.)X#p. . ..... (33) 



Expressions (31) and (32) for the isothermal compressor are especially useful 
is standards of comparison for the economy of the compressors using methods 
3ther than isothermal. It will be found that the work per cubic foot of either 
ow pressure or cooled high-pressure gas is less by the isothermal process than 
3y any other process discussed later, and that it is the limiting case for the 
economy of multi-stage compressors with a great number of stages. The fact 
that this process of isothermal compression is seldom if ever approached in 
practice does not make it any the less a suitable basis for comparison. 

Example 1. Method of calculating Diagram Fig. 24. 
Assumed Data. 

p a =p b =, 2116 Ibs. per square foot. V a = V d =0. 

P c =Pd = 18,000 Ibs. per square foot. Capacity =5 cu.ft. 



To obtain point C, 



Pb 5X2116 
PC b ~ 18,000 

.59, PC = 18,000. 



84 ENGINEERING THERMODYNAMICS 






Intermediate points B to C are obtained by assuming various pressures and findir 
corresponding volumes as for V c . 

Example 2. To compress and deliver 5 cu.ft. of air from atmospheric pressure (211 
Ibs. per square foot) to 8.5 atmospheres (18,000 Ibs. per square foot) isothermal] 
without clearance, how much work is necessary? 

P 6 =2116 P c = 18,000 

F 6 =5 

-- 8 ^ 17 CiSft 

Vc~Pb d 

Work of admission =P b Vi> =2116 X5 = 10,585 ft.-lbs. 

p 
Work of compression =P b V b log e ~ =10,585 Xlog e 8.5 =22,600 ft.lbs. 

f b 

Work of delivery =P C V C = 10,585 ft.-lbs. 

Total work = 10,585 + 22,600 - 10,585 = 22,600 ft.-lbs. 

Or by the general formula, 

TF = (sup.pr.)(L.P.Cap.) log e R p =2116x5 xlog e 8.5 =2116x5x2.14 =22,652 ft.-lb, 

Prob. 1. How many cubic feet of free air may be compressed and delivered pc 
minute from 14 Ibs. absolute to 80 Ibs. per square inc 11 , absolute per horse-power in 
compressor with zero clearance if compression is isothermal? 

Prob. 2. Gas is being forced through mains at the rate of 10,000 cu.ft. per minul 
under a pressure of 5 Ibs. per square inch above atmosphere. The gas is taken into tt 
compressor at atmospheric pressure and compression is isothermal. What hors< 
power will be needed at sea level and at an elevation of 5000 feet? 

Prob. 3. Natural gas is drawn from a well, compressed isothermally and forced throug 
a main at the rate of 200,000 cu.ft. per hour measured at the pressure on the suctio 
side. What steam horse-power will be required to operate the compressor if th 
mechanical efficiency be 80 per cent? Suction pressure is 8 Ibs. per square inc 
absolute, delivery pressure 60 Ibs. per square inch absolute. 

Prob. 4. A vacuum cleaning pump is required to maintain a pressure of 14 Ibs. p( 
square inch absolute, move 500 cu.ft. of free air per minute and discharge it againt 
an atmospheric pressure of 15 Ibs. per square inch absolute. What horse-power wi 
be required (isothermal)? 

Prob. 5. A blower furnishes 45 cu.ft. of ir a minute at a pressure of 5 ins. of mercur 
above atmosphere. Assuming compression to be isothermal and supply pressure to t 
atmospheric, what horse-power will be needed? 

Prob. 6. A compressor has a piston displacement of 3 cu.ft. At what speed can it I 
run if air be compressed isothermally from 1 to 10 atmospheres and the horse-pow< 
supplied is 100? 

Prob. 7. A tank of 1000 cu.ft. capacity contains air at atmospheric pressure. . 
compressor taking air from atmosphere compresses it isothermally and discharges it inl 
the tank until the pressure reaches 100 Ibs. per square inch gage. What horse-pow( 
will be required to fill tank at this pressure in ten minutes? 

Prob. 8. A compressor receives air at atmosphere and compresses it isothermally i 
five atmospheres above atmosphere. It takes in 1000 cu.ft. of free air per minut* 
How much would the capacity increase if the discharge pressure dropped to 3 atmo 
pheres and the horse-power remained the. same? 



WORK OF COMPRESSORS 



85 



Prob. 9. Suppose that the pressure in the above problem were raised to 8 atmos- 
Ueres. How much would the capacity decrease if the horse-power remained the same 
id how much more power would be required to keep the capacity the same? 

Prob. 10. By means of suitable apparatus, the water from the side of a waterfall is 

verted to a vertical shaft, and in falling 126 ft. compresses air from atmospheric pressure 

a value equal to 90 per cent of the head of the water. To deliver 1000 cu.ft. of com- 

essed air per hour, how much water is required if the work of falling water is 80 per 

;;nt useful in compressing the air? 

4. Single-stage Compressor with Clearance, Isothermal Compression, 
2ycle 2) . Work, Capacity, and Work per Cubic Foot in Terms of Pressures 
ad Volumes. 




FIG. 25. One-Stage Compressor Cycle 2, Clearance, Isothermal. 



Referring to Fig. 25, the work of the cycle i^by areas. 
Net work 



-HADF--EBAH 

= Area ABC D. . 

[t is easily seen that this area is also equal to (JBCL) (JADL), both 
which are areas of the form evaluated in the preceding section- Accordingly 



Net work 



- JADL, 



86 ENGINEERING THERMODYNAMICS 

Algebraically, 



(3 



which is the general expression for the work of the cycle in foot-pounds whi 
pressures are in pounds per square foot, and volumes in cubic feet. Substituti] 
the symbolic equivalents and using pressures in pounds per square inch, the 
results, since (F& F a ) = (L. P. Cap.), 

Work = 144 (sup.pr.)(L. P. Cap.) log e R p , .... (3 

which is identical with Eq. (30), showing that for a given low-pressure capaci 
the work of isothermal compressors is independent of clearance. The value 
the low-pressure capacity (F& V a ) may not be known directly, but m; 
be found if the volume before compression, F&, the clearance volume befc 
re-expansion, Vd, and the ratio of delivery to supply pressure, R p , are know 
thus 

V a =V d R p) 

from which 

(L. P. Cap.) = (VV- Ftffl,) ......... (2 



From Eq. (35) the work per cubic foot of low-pressure gas is, in foot-pounds, 

W 



144(sup.pr.) 



(L.P. Cap.) 

and the work per cubic foot of high-pressure gas delivered, ft.-lbs. 

W 



(H.P. Cap.) 



144(sup.pr.)# p log e R 



By comparison, Eqs. (37) and (38) are found to be identical with (31) a 
(32) respectively, since clearance has, as found above, no effect on the work del 
for a given volume of gas admitted, however much it may affect the work of i 
cycle between given volume limits or work per unit of displacement. 

It is interesting to note that the work areas of Figs. 24 and 25 are eqi 
when plotted on equal admission lines AB or delivery lines CD and a|j 
horizontal intercept xy will be equal in length on both if drawn at thesa:f 
pressure. 

In what precedes, it has been assumed that AB represents admission voluii 
and CD represents delivery volume which is true for these established eye I 



WORK OF COMPRESSORS 87 

sference, but it is well to repeat that for real compressors these are only 
>parent admission and delivery lines, as both neglect heating and cooling 
.fects on the gas during its passage into and out of the cylinder. Also that 
: real compressors the pressure of the admission line cannot ever be as high 
: T, the pressure from which the charge is drawn and the delivery pressure must 
jj necessarily higher than that which receives the discharge, in which cases 
,ie volume of gas admitted, as represented by AB, even if the temperature 
id not change, would not equal the volume taken from the external 
ijipply, because it would exist in the cylinder at a lower pressure than it 
,'iginally had, and a similar statement would be true for delivered gas. 

! Problems. Repeat all the problems of the last section, assuming any numerical 
jilue for the clearance up to 10 per cent of the displacement. 

i 5. Single-stage Compressor Isothermal Compression. Capacity, Volu- 
metric Efficiency, Work, Mean Effective Pressure, Horse-power and Horse- 
ower per Cubic Foot of Substance, in Terms of Dimensions of Cylinder and 
learance. 

Consider first the case where clearance is not zero. Then Fig. 25 is the 
;jference diagram. 

Let D = displacement = volume, in cubic feet, displaced by piston in one 

stroke = area of piston in sq.ft. X stroke in ft. = (F& V d ). 
" (H. P. Cap.) = high pressure capacity = vol. cu.ft. of gas delivered per 

cycle at temperature equal to that of supply =(F C V d ); 
(L. P. Cap.) = low pressure capacity = vol. in cu.ft. of gas entering 
cylinder per cycle =(F& F a ); 

L.P. Cap. V b -V a 
E v = volumetric efficiency = ^j = y _y > 

" Cl = volume of clearance, cubic feet= V d 

" c = clearance volume expressed as a fraction of the displacement; 

Cl V 
=~. = v * whence Cl = cD', 

LJ V b v a 

W 
M.E.P. =mean effective pressure, Ibs. per square foot = yr-; 

W 

m.e.p. = mean effective pressure, Ibs. per square inch = ~ ; 

" N number of revolutions per minute; 

" n number of cycles per minute; 

N 
z = number of revolutions per cycle = ; 

" I.H.P. = indicated horse-power of compressor; 



88 ENGINEERING THERMODYNAMICS 






The low-pressure capacity of the single-stage isothermal compressor wit 
clearance is, 



but 






Whence (L. P. Cap.) = ( V b V d j/ j for which may be substituted the symbo 
for displacement and clearance volumes, thus 



(L. P. Csip.)=D+cD-cDR p , 
= D(\+c-cR p ) 



For convenience the term, Volumetric Efficiency, E v is introduced. Sim 
this is defined as the ratio of the low-pressure capacity to the displacement, 

(K P. <*>.)_, 



Referring to Eq. (35) it is seen that the value of (L. P. Cap.) can 1 
substituted from Eq. (39) and the result is: 

Work per cycle, foot-pounds, in terms of supply pressure, pound spersq ua 
inch, displacement cubic feet, clearance as a fraction of displacement, and rat 
of delivery to supply pressure is, 

HF = 144 (sup.pr.)D(l+c-cRp)\ogeR 

or in the terms of the same quantities omitting clearance and introduce 
volumetric efficiency, E v , 

/ TF=144(sup.pr.) DE v \og e R P , . . (4 

To obtain the mean effective pressure for the cycle, the work done per cy< 
is divided by displacement, D. 

Mean effective pressure, pounds per square inch, 

W 

(m.e.p.) = 



144D' 
whence 

(m.e.'p.) = (sup.pr.)(l+c cR P ) \og e R P 

or 

(m.e.p.) = (sup.pr.)^,, log c R p * . ( z 






WORK OF COMPRESSORS 89 



The indicated horse-power of the isothermal compressor is equal to the 
work per minute, in ft.-lbs. divided by 33,000. If n cycles are performed 
per minute, then 



(45) 
(46) 



^Introducing the effective area of the piston, in square inches, a, and the piston 
Jspeed S, feet per minute, then since 



44 2z 

(sup.pr> , 
LILR - 66,000 7^' logejRp ' ' ...... >.:' 

The same expression for the indicated horse-power may be derived by the 
.substitution of the value of (m.e.p.) Eq. (44) in the following general expres- 
ision for indicated horse-power. 

= (m.e.p.)o 
33000 X 2z 

Example 1. Method of calculating Diagram Fig. 25. 
Assumed Data: 

P a =Pb =2116 Ibs. per square foot; 
P d =P c =18,000 Ibs. per square foot; 
c =3 per cent. L.P. Cap. =5 cu.ft. s = 1. 

To obtain point D: 

From formula Eq. (39), L.P. Cap. =D(l+c-cRp) or 5=Z)(l+.03-.03x8.5), 
or 

D =6.5 cu.ft. and Cl. =.03 X6.5 =.195 or approximately .2 cu.ft. 

/. V d = .2 cu.ft., P d = 18,000 Ibs. sq.ft. 

To obtain point A : 

PaV a =P d V d or 7 a =!^ = 8.5X.2 = 1.7, 

* a 

.'. F a = 1.7 cu.ft., P a =21161bs. sq.ft. 



90 



ENGINEERING THERMODYNAMICS 



Intermediate points D to A are obtained by assuming various pressures and finding the 
corresponding volumes as for F a . 
To obtain point B: 



=6.7 cu.ft. 



= 2116 Ibs. sq.ft. 



To obtain point C: 



':. V c = .79 cu.ft., 



PC = 18,000 Ibs. sq.ft. 



Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to 
atmospheres isothermally in a compressor having 4 per cent clearance. What mi 
be the displacement work, per 100 cu.ft. of supplied and delivered air, and horse-powe]; 
of machine? Speed is 150 R. P.M., compressor is double acting and stroke = 1.5 diameters 
Neglect piston rods. 



D = (L. P. Cap.) -;-#, and # = (!+. 04 -.04x8.5) =.7, 
.'. D = 1000 4- .7 = 1428 cu.ft. per minute. 

Work per cu.ft. of supplied air = (sup.pr.) 144 log e R p = 144 X 14.7 log e R P =4530 ft.-lbs.; 
Hence the work per 100 cu.ft. =453,000 ft.-lbs. 

Work per cu.ft. of delivered air = 144(sup.pr.)# p log e R p = 144 X14.7 X8.5 X2.14 =38,^ 

ft.-lbs. 

Hence the work per 100 cu.ft. =3,855,000 ft.-lbs. 



D = 



33000 
1428 



150X2 



= 4.76 cu.ft. per stroke. 



Hence cylinder diameter = 



= 1.59 feet = 19.1 inches. 



Prob. 1. How many cubic feet of free air per minute may be compressed isotl 
mally to 100 Ibs. per square inch absolute in a compressor having 6 per cent clean 
if the horse-power supplied is 60? 

Prob. 2. A compressor has a cylinder 18x24 ins., clearance 4 per cent, is doul 
acting and runs at 150 R.P.M. If it compresses air from atmosphere to 100 Ibs. 
square inch gage, what will be its high- and low-pressure capacity, its horse-power, 



WORK OF COMPRESSORS 91 

fiow will the horse-power and the capacity compare with these quantities in a 
V hypothetical compressor of the same size but having zero clearance? How will 
;he horse-power per cubic foot of delivered air compare? 

Prob. 3. A manufacturer gives for a 10|xl2 in. double-acting compressor running 

it 160 R.P.M. a capacity of 177 cu.ft. of free air per minute for pressures of 50-100 Ibs. 

3er square inch gage. What clearance does this assume for the lowest and highest 

pressure if the compression is isothermal? The horse-power is given as from 23 to 35. 

r'dheck this. 

Prob. 4. Air enters a compressor cylinder at 5 Ibs. per square inch absolute and is 
compressed to atmosphere (barometer = 30| ins.). Another compressor receives air 
rat atmosphere and compresses it to 3 atmospheres. If each has 6 per cent clearance 
,:what must be the size of each to compress 1000 cu.ft. of free air per minute, how will 
ijthe total work compare in each machine, and how will the work per cubic foot of high 
;jand low pressure air compare in each? Assume compression to be isothermal. 

Prob. 5. 1800 cu.ft. of free air per minute is to be compressed isothermally to apres- 
jsure of 100 Ibs. per square inch gage. What must be the displacement and horse-power 
of a hypothetical zero clearance compressor, and how will they compare with those of a 
i compressor with 6 per cent clearance? 

Prob. 6. Consider a case of a compressor compressing air isothermally from atmos- 
phere to 100 Ibs. per square inch gage. Plot curves showing how displacement and 
horse-power will vary with clearance for a 1000 cu.ft. free air per minute capacity 
taking clearances from 1 per cent to 10 per cent. 

Prob. 7. Two compressors of the same displacement^ namely 1000 cu.ft. per min- 
ute, compress air isothermally from 50 Ibs. per square inch gage to 150 Ibs. per square 
inch gage. One has 5 per cent clearance, the other 10 per cent. How will their capaci- 
ties and horse-power compare with each other and with a no clearance compressor? 

Prob. 8. A 9 Xl2 in. compressor is compressing air from atmosphere to 50 Ibs. gage. 
How much free air will it draw in per stroke, and how much compressed air will it dis- 
charge per stroke for each per cent clearance? 

Prob. 9. The volumetric efficiency of a compressor is .95 as found from the indicator 
card. It is double acting and has a cylinder 18 X 24 ins. What will be its capacity and 
required horse-power for 100 Ibs. per square inch gage delivery pressure? 

6. Single-stage Compressor, No Clearance Exponential Compression, 
(Cycle 3) . Work, Capacity and Work per Cubic Foot, in Terms of Pressures 
and Volumes. The cycle of the single-stage exponential compressor without 
clearance is represented by Fig. 26. Referring to work areas on this diagram, 

Net work A BCD = compression work EBCG 
+ deli very work GCDF 
admission work FARE. 

Algebraically, 

s 1 



92 

But 

or 

and 



ENGINEERING THERMODYNAMICS 



P V T7 *-l p V V s 1 
* ' c V c * J ' 6 ' & > 



-l] 



18000 
14400 


(H.P. Cap 1 
* Cold) *'(H.P. Cap, 






Hot) j 










, - 




















































































































































































































































D 








| 
































































































































































I 












































































m ;00 

^(Sup.Pr^ 
























































































































































































i 






















































































1 






















































































!| 






















































































: 


\ 
















































































































































































1 








f 






















































































\ 


















































































1 






3 


















































































\ 
























































































\ 






\ 


















































































\ 






\ 




























































































































































Pressure? in Pounds per Square 


















\ 







\ 










































































- 






























































































& 

V- 
















































































X 






\ 
















































































5 






^ 


s 






















































































\"< 














































































^ 








\ 


(S 






















































































k V. 














































































$ 








\ 


& 














































































Sa 
























































































A^ 








\ 














































































X< <X 










^ 












































































\ 


'? 










s 


























































































\ 


























































































\ 












































































\ 


























































































w 












s 














































































\ 












\ 


























































































V 




























































































^ 














































































X 












s 
















































































s 












^ 
















































































-.^ 










*s. 


































- 
















































^, 










"V 


s. 
















































































-*^, 






























































































^ 






^^ 




--^. 


^ 


















































































**. 


^^ 


^: 




s^ 

























































































































































s 




=H= 


































































































































































































































































































1 ? 3 4 5 
* Volumes in Cubic Feet 



1 



L.P. 



FIG. 26. One-Stage Compressor Cycle 3, No Clearance, Exponential. 
Substituting above 






WORK OF COMPRESSORS 93 

Thence 

<**> 



iCq. (48) gives the work in foot-pounds for the execution of the cycle when 
)ressures are in pounds per square foot, and volumes in cubic feet. 
The equivalent expression for pressures in pounds per square inch is 

s 1 

(49) 



When there is no clearance, as before, F& represents the entire vol- 

}ime of displacement, which is also here equal to the volume admitted 

. 

!L. P. Cap.), p b is the supply pressure (sup.pr.) pounds per square inch 
iibsolute and is the ratio of delivery to supply pressure, R p . 

Accordingly, the work of an exponential, single-stage compressor with no 
Clearance is 

s-l 

Tf = 144^ T (sup.pr.)(L.P.Cap.)(^/ -l) (50) 

S L \ / 

The work per cubic feet of low pressure gas, foot-pounds is 

8-1 

W s ( ~T~ \ 

/T p ^ v 144 -(sup.pr.) I R P 1 (51) 

(L. P. Cap.) s 1 \ 

n 

Before obtaining the work per cubic foot of high-pressure gas, it is neces- 
ary to describe two conditions that may exist. Since the exponential com- 
jiression is not isothermal, it may be concluded that a change in temperature 
fvill take place during compression. This change is a rise in temperature, 
jind its law of variation will be presented in another chapter. 

1. If the compressed air is to be used immediately, before cooling takes 
i)lace, the high-pressure capacity or capacity of delivery will be equal to the 
volume at C, V c and may be represented by (H. P. Cap. hot). 

2. It more commonly occurs that the gas passes to a constant-pressure 
lolder or reservoir, in which it stands long enough for it to cool approximately 
o the original temperature before compression, and the volume available 
liter this cooling takes place is less than the actual volume discharged from 
he cylinder in the heated condition. Let this volume of discharge when 
educed to the initial temperature be represented by (H. P. Cap. cold) which 
s represented by Vt, Fig. 26. 



94 ENGINEERING THERMODYNAMICS 

Since B and C in Fig. 26 lie on the exponential compression line, P&TV 



or 

(L. P. Cap.) = (H. P. Cap. hot) (R,)? (52 

Hence, the work in foot-pounds per cubic foot of hot gas delivered fron 1 
compressor is 

W s i / s-i 

/~r~r -pj ^r~^ V~ TV *""~ -*- i A T yoLlJ3JJi * J f\jp ^ \ **'' p ^ 

(ti. r. uap. notj s i \ 

On the other hand, B and K lie on an isothermal and 
since P* = P c , 



whence 

(L. P. Cap.) = (H. P. Cap. cold)# p (54 

The work foot-pounds per cubic foot of gas cooled to its original temper* 
ture is, therefore, 

W s / tzl \ 

= 144 -(sup.pr.)R p (R p -ij, . . . . (5f 



(H. P. Cap. cold) s-l 
or 

W 



. . . . (5H 



This last equation is useful in determining the work required for the storir 
or supplying of a given amount of cool compressed air or gas, under conditioi 
quite comparable with those of common practice. 

Example 1. Method of calculating Diagram Fig. 26. 
Assumed Data: 

p a =p b =2116 Ibs. per square foot; P c =Pa= 18,000 Ibs. per square foot. 

Cl =0; V a = V d =0; L. P. Capacity =5 cu.ft.; s = 1.4 (adiabatic value of ; J 

To obtain point C: 



or 



_ 
P 6 /P & =8.5; Iog e 8.5 = .929, and .71 log* 8.5 = .665; (P c /P 6 )i-4 =4.6, 



WORK OF COMPRESSORS 95 

lence V c = 5 -J-4.6 = 1.09 cu. ft. P c = 18,000 Ibs. per sq.ft. 

.ntermediate points B to C are obtained by assuming various pressures and finding 
,he corresponding volumes as for V c . 



Example 2. To compress 5 cu.ft.of air from atmospheric pressure (2116 Ibs. per 
quare foot) to 8.5 atmospheres (18,000 Ibs. per square foot) adiabatically and with no 
learance requires how many foot-pounds of work? 

P b =2116 Ibs. sq.ft.,, P c = 18,000 Ibs. sq.ft., 

F 6 =5 cu.ft. 

F C =T]JT n =5-4.57 = 1.092 cu.ft. 



^ork of admission is 

P b V =2116X5 = 10,585 ft.-lbs. 

Work of compression, using y to represent the adiabatic value of s is, 



Work of delivery is 

P C V C = 18,000 X 1.092 = 19,650 ft.-lbs. 

:otal work = 19,650+22,350-10,585=31,425 ft.-lbs., 

>y the formula Eq. (50) directly 

/ -1 \ 
TT = 144^j(sup. pr.) (L. P. Cap.) \R p r -l.j 

= 144+3.46 X2116 X5 X[(8.5)' 29 -l]; 

= 144+3.46 X2116 X5 X.86 =31,450 ft.-lbs. 

rob. 1. A single-stage zero clearance compressor compresses air adiabatically from 
6 atmospheres. How many cubic feet of free air per minute can be handled if the 
iompressor is supplied with 25 H.P. net? 

Prob. 2. The same compressor is used for superheated ammonia under the same 
pressure conditions. For the same horse-power will the capacity be greater or less 
ad how much? 

Prob. 3. A dry-vacuum pump receives air at 28 ins. of mercury vacuum and delivers 
; against atmospheric pressure. What will be the work per cubic foot of low-pressure 
ir and per cubic foot of high-pressure air hot? Barometer reads 29.9 ins. 



96 ENGINEERING THERMODYNAMICS 

Prob. 4. The manufacturer gives for a 10iXl2 in. double acting compressor run 
ning at 160 R.P.M., a capacity of 177 cu.ft. of free air per minute and a horse-power o 
25 to 35 when delivering against pressures from 50 to 100 Ibs. Check these figures. 

Prob. 5. A set of drills, hoists, etc., are operated on compressed air. For their opera 
tion 3000 cu.ft. of air at 70 Ibs. gage pressure are required per minute. What must br 
the piston displacement and horse-power of a compressor plant to supply this air i 
compression is adiabatic and there is assumed to be no clearance? 

Prob. 6. Air is compressed from atmosphere to 60 Ibs. per square inch gage by iji 
compressor having a 12x18 in. cylinder and running at 100 R.P.M. Find its capacity 
and horse-power at sea level and loss in capacity and horse-power if operated at an alti 
tude of 10,000 ft. for zero clearance. 

Prob. 7. 10,000 cu.ft. of free air per minute at a pressure of 15 Ibs. above atmos,i 
phere are compressed and delivered by a blowing engine. Find the horse-power requires 1 
to do this and find how much free air could be delivered by same horse-power if th|j 
pressure were tripled. 

Prob. 8. In a gas engine the mixture of air and gas is compressed in the cylinde 
before ignition. If the original pressure is 14 Ibs. per square inch absolute, fim 
pressure 85 Ibs. absolute and compression is adiabatic, what will be the work c 
compression only, per pound of mixture? 

NOTE: Weight per cubic foot may be taken as .07 and y as 1.38. 

Prob. 9. A vacuum pump is maintaining a 25-in. vacuum and discharging the a 
removed against atmospheric pressure. Compare the work per cubic foot of low pre: 
sure air with that of a compressor compressing from atmosphere to 110 Ibs. above atmo: 
phere. 

7. Single-stage Compressor with Clearance, Exponential Compressioi 
(Cycle 4). Work, Capacity, and Work per Cubic Foot in Terms of Pressure 
and Volumes. When clearance exists in the cylinder, it is evident that 
volume equal to the clearance, V*, will not be expelled during the deliver 
of compressed gas, and -this volume will expand with fall in pressure as tl 
piston returns, causing pressure-volume changes represented by the line DJ 
on the diagram, Fig. 27. Until the pressure has fallen to that of supply, tl! 
admission valve will not open, so that while the total volume in the cylind< 
at end of admission , is 7 6 , the volume V a was already present by reasoj 
of the clearance, and the volume taken in is (F& Va) which is the low-pressui* 
capacity (L. P. Cap.). 

The work area of the diagram is A BCD, which may be expressed as 

Work area = JBCL- JADL, 

which areas are of the form evaluated in Section 6. Hence, the above expressic 
in algebraic terms is 



WORK OF COMPRESSORS 



97 



is is the general expression for the work of the cycle, in foot-pounds, when 
the pressures are expressed in pounds per square foot, and volumes in cubic 
feet. Using symbolic equivalents 



(H.P.'Cap.Hot) 



-1, .... (58) 



18000 




234 

Volumes in Cubic Feet 
U.P. Cap.) 



D 



FIG. 27. One-Stage Compressor Cycle 4, Clearance, Exponential. 

(58) is identical with Eq. (50), showing that for adiabatic as for 
lermal compressors, the worl done for a given low-pressure capacity is inde- 
pendent of clearance. Due to this fact, the expressions derived for the expo- 
nential compressor without clearance will hold for that with clearance: 



Work, in foot-pounds per cubic foot of low-pressure gas is, 

W ' s 



(L.P.Cap.r s -^i-' '" 
Work, in foot-pounds per cubic foot of hot gas delivered is, 



(59) 



(H.P. 



8-1 

s~ 



98 ENGINEERING THERMODYNAMICS 

Work, in foot-pounds per cubic foot of cooled gas to its original temperature is, 

= 144-^-(del.pr.)(W -l] . (61) 



(H. P. Cap. cold) s-l 

The relation of high-pressure capacity either hot or cold to the low-pressure 
capacity is also as given for the case of no clearance, as will be shown. 

In Fig. 27, the high-pressure capacity, hot, is DC=V c V d . The low- 
_ 11 11 

pressure capacity is AB= V b V a , but VcP c * = V b Pb s and TW = TW, or 
V b = VcR^ and V a = Va 



i 
Hence (L. P. Cap.) = (H. P. Cap. hot)tf/ ...... (62) 

If the delivered gas be cooled to its original temperature, then the volume 
after delivery and cooling will be 

. (L. P. Cap.) 
(H. P. Cap. cold) =- -- 5 - , 

Lip 

or 

(L.P. Cap.) = (H. P. Cap. cold)R p .... (63) 

From the work relations given above, it is seen that in general, the work 
per unit of gas, or the horse-power per unit of gas per minute is independent 
of clearance. 

8. Single-stage Compressor Exponential Compressor. Relation between 
Capacity, Volumetric Efficiency, Work, Mean Effective Pressure, Horse- 
power and H.P. per Cubic Foot of Substance and the Dimensions of Cylinder 
and Clearance. As indicated on Fig. 27, for the single-stage exponential com- 
pressor with clearance, the cylinder displacement D, is (Vb Vd). The low- 
pressure capacity per cycle is (L. P. Cap.) = (F& F a ). The actual volume of 
gas or vapor delivered by the compressor is (H. P. Cap. hot) = (V c ~ Vd). 
This is, in the case of a gas at a higher temperature than during supply, 
but if Qooled to the temperature which existed at B will become a less volume. 
This delivered volume after cooling is symbolized by (H. P. Cap. cold) and is 

equal to (L. P. Cap.) Xy -r or '- =- '- where R p is the ratio of delivery 
(del.pr.) Hp 

pressure to supply pressure. 

Volumetric efficiency, E v , already defined as the ratio of low-pressure 
capacity to displacement is 

n-y a _(L.P.Cap.) 
*"-V^-V d - ~D ' 






Cl 



WORK OF COMPRESSORS 99 



earance, c, expressed as a fraction of the displacement is the ratio of 
irance volume, Cl, to displacement, D, and is, 

Cl__ V d 
= D'V b -V d 

iean effective pressure, pounds per square foot (M.E.P.), is the mean height 
Oithe diagram or the work area W, divided by displacement, D. If expressed 
abounds per square inch the mean effective pressure will be indicated by 



|| Let (I.H.P.) be indicated horse-power of the compressor; 
N the number of revolutions per minute; 
n the number of cycles per minute and 
z the number of revolutions per cycle, whence nXz = 

I Then, the low-pressure capacity is 

(L.P. 
U 



3ice the re-expansion DA is exponential and similar to compression as to 
ilue of s, whence j 

(L. P. Cap.) = (Vt,-Va) = V b - V d R p ; 



i 
D+cD-cDR P * ; 



(L. P. Cap.)=-D(l+c-c3) (64) 

. 
*om this, by definition, the volumetric efficiency is 

(L.P. Cap.) i f . 

J% = 7: =l+c cR p * (65) 



Referring to Eq.(57), in which may be substituted the value Eq. (64) for 
7 & F a ), the work of the single-stage exponential compressor in terms of dis- 



100 ENGINEEEING THERMODYNAMICS 

placement, clearance (as a fraction of displacement), and pressures of suj 
and delivery in pounds per square foot is, 



or using pressures, pounds per square inch, and inserting the symbols, thi 
may be stated in either of the following forms: 



w= 



-c/^) \Rp - -l|. . . . 




The mean effective pressure in pounds per square foot is this work dividfj 
by the displacement, in cubic feet, and may be converted to pounds per squa 
inch by dividing by 144, whence 

Mean effective pressure, pounds per square inch. 

6 






(7 



The indicated horse-power of the single-stage exponential compressor frcjl 
(67) is, 

T H P - Wn . s (sup.pr.)nDE v [ ^_ 1 ( 

LKR -33000"^! ~ ~^2&r 

Where n is the number of cycles per minute, or in terms of piston speed S a I 
effective area of piston, square inches, and z the number of revolutions 
cycle, i 

s (sup.pr.)qff p I" i^- 1 1 
[ - 66()()02 ^ 






Since it was found in Section 7, that the work per unit volume of gas is t 
same with clearance as without clearance, the horse-power per cubic foot i 
minute will also be independent of clearance. (See Eqs. (51), (53) and (56) ).j 

Horse-power per cubic foot of gas supplied per minute 

I-H.P. _!_fouprOU ^J 
n(L.P.Cap.)~-l 229:2 Kv 



WOKK OF COMPRESSORS, ' 101 

The horse-power per cubic foot of hot gas delivered per minute is 

LH.P. (sup.pr.) I f =! 1 

n(H.P.Cap.hot) = ^I 229.2 R "' * * 

brse-power per cubic foot of gas delivered and cooled is 

I-H.P. _ __ (sup.pr.) T til _ I 

n(H.P. Cap .cold) ~s-l 229.2 ^[/^ ' l \' ' 

s (del.pr.Ho ,1 
= . 122932 I*' 8 ' ' < 76 > 

i In the above formulae (del.pr.) and (sup.pr.) indicate delivery pressure and 
spply pressure, in pounds per square inch. 

Example 1. Method of calculating Diagram, Fig. 27. 
\3sumed data: 

p a =p b =2116 Ibs. per square foot. 

p c =p d = 18,000 Ibs. per square foot. 

Cl. =3.5 per cent. L. P. Capacity =5 cu.ft. s = 1.4. 
'3 obtain point D: 

/ i \ / \ .715 

L. P. Ca,p.=D\l+c-cR7) or 5=D U +.035 -.035 (8.5) / ; 



ence 



D-5^(1+.035-.035X4.6) =5.72 cu.ft. and C7 = . 035x5.72 =.2 cu.ft 
.*. V d = .2 cu.f t. ; P d = 18,000 Ibs. sq.ft. 



jo obtain point A : 



= 4.6X.2=.92; 
/. Fa =.92 cu.ft.; P a =2116 Ibs. sq.ft. 

itermediate points D to A are obtained by assuming various pressures arid finding the 
^responding volumes as for V a . 

o obtain point B: 

. P. Cap. =.92 +5 =5.92. 



/. F & = 5.92 cu.ft.; P b =2116 Ibs. sq.ft. 



ENGINEERING THERMODYNAMICS 



To obtain point C: 



= 5.92 -=-4.6 = 1.29 cu.ft. 
.'. V c = 1 .29 cu.ft. ; PC = 18,000 Ibs. sq.ft. 

Example 2. It is required to compress 1000 cu.ft. of air per minute from 1 to ': 
atmospheres absolute so that s = 1.4, in a compressor having 4 per cent clearance. Wl t 
must be the displacement of the compressor, work per 100 cu.ft. of supplied and deliver i 
air, hot and cold, and horse-power of machine? Speed is 150 R.P.M., compressor- 
double acting and stroke = 1.5 diameters. 



D =L. P. Cap. +E V , and E v = (l -f c -cR^ ) . 
.*. # p =(l+.04-.04x(8.5)- 71 ) = .86; 
/. D = 1000 + .86 = 1 162 cu.ft. per min. 



.s-l 



Work per cubic foot of supplied air = 144 -- - (sup.pr)[/tV s 1], 

s 1 

= 144 X3.46X 14.7 X. 86 =6300 ft.-lbs. 
.'. Work per 1000 cu.ft. = 6,300,000 ft.-lbs. 



Work per cubic foot of delivered air cold is R p times work per cubic foot of suppliec 
hence work per 100 cu.ft. of delivered cooled air is 5,350,000 ft.-lbs. 

Work per cubic foot of delivered air hot is R p ~^ times work per cubic foot of supplied 
hence work per 100 cu.ft. of hot delivered air is 2,800,000 ft.-lbs. 



or 



(m.e.p.) = 3.46 X 14.7 X. 86 X. 86 =37.7 Ibs. per square inch. 



.'. ^=5690 or d = 17.85. 
a =250 sq.inches. S =670 ft. per min. 



.'. I.H.P.=191. 



WORK OF COMPRESSORS io< 

Prob. 1. A dense-air ice machine requires that 4000 cu.ft. of air at 50 Ibs. per squar< 
inch absolute be compressed each minute to 150 Ibs. per square inch absolute. The 
compression being such that s = 1.4, clearance being 6 per cent, find the work required 
What would be the work if clearance were double? Half? 

Prob. 2. The compressor for an ammonia machine compresses from one atmos- 
phere to 8 atmospheres absolute. With adiabatic compression and 4 per cent clear- 
ance, what will be work per cubic foot of vapor at the low pressure and at the high! 
Assume vapor to be superheated. 

Prob. 3. On a locomotive an air-brake pump compresses air adiabatically frocc 
atmosphere to 80 Ibs. per square inch gage. It is required to compress 50 cu.ft. of free 
air per minute and clearance is 5 per cent. What horse-power must be supplied to it! 

Prob. 4. In a manufacturing process a tank must be maintained with a vacuum oi 
29 ins. when barometer reads 30 ins. To do this 100 cu.ft. of carbon dioxide must be 
removed from it per minute and returned under atmospheric pressure. Compression is 
adiabatic and clearance 7 per cent. How much power must be supplied to compressor 
and what should be its displacement? 

Prob. 5. Two compressors each 12x18 in., double acting, with 8 per cent clear- 
ance, and running at 150 R.P.M. compress in the one case air, in the other carbon disul- 
phide. The compression being adiabatic in each case, what (a), is the difference in 
power required, (6), in low-pressure capacities? Take pressures as 2 and 15 atmos- 
pheres of 26 inches mercury. 

Prob. 6. A compressor is supplied with 40 horse-power. If it draws in air from 
atmosphere to what pressure can if 500 cu.ft. per minute be compressed, when s = 1.38 
and clearance 10 per cent? 

Prob. 7. For forcing gas through a main, a pressure of 50 Ibs. per square inch gage 
is required. W'hat is the work done per cubic foot of high-pressure gas, if a compressor 
having 6 per cent clearance is used, and s for the gas is 1.36? What should be its dis- 
placement? 

Prob. 8. A gas compressor 20x22 ins. has a volumetric efficiency of 90 per cent, 
supply pressure =4 Ibs. per square inch and delivery 110 Ibs. per square inch gage. 
W 7 hat are its L. P., H. P., hot and cold capacities, and its I.H.P. if single acting at 70 
R.P.M. when s = 1.35? 

9. Two-Stage Compressor, no Clearance, Perfect Intercooling, Exponential 
Compression, Best Receiver Pressure, Equality of Stages (Cycle V). Work and 
Capacity in Terms of Pressures and Volumes. The common assumption in con- 
sidering the multi-stage compressor is that in passing from one cylinder to the 
next, the gas is cooled to the temperature it had before entering the com- 
pressor, which has already (Section 2), been defined as " perfect inter cooling." 

This condition may be stated in other words by saying that the product 
of pressure and volume must be the same for gas entering each cylinder. If 
then the volume and pressure of gas entering the first stage be determined, 
fixing the volume entering the second stage will determine the pressure of the 
gas entering the second stage, or fixing the pressure of the gas entering the 
second stage will determine the volume that must be taken in. 

Using subscripts referring to Fig. 28, for the no clearance case," 

P b V, = P d V d (77) 



104 ENGINEERING THERMODYNAMICS 



The net work of the compressor, area ABCDEF = area ABCH first stage 
area HDEF second stage. Using the general expression, Eq. (48) for these 
work areas with appropriate changes in subscripts 



8-1 

' 



stage) 

S-l 



+^^ p dVd\(~J -1 . . (second stage) 
But from the above, and since P c = P<i, 



which is the general expression for work of a two-stage compressor without 
clearance, perfect intercooling, and may be restated with the usual symbols as | 
follows : 



W = 144~ T (sup.pr.)(L. P. Cap.) (R P i)-+(R p2 y-2 , . (79) 

L J 

n which (R P i) and R P z) are the ratios of delivery to supply pressures for the j 
irst stage and for the second stage respectively. From Eq. (79), work per i 
jubic foot of gas supplied is, 

W s f s ~ 1 s ~ 1 1 

"~~ 2 ' ' (80) 



Work per cubic foot of gas discharge and cooled to its original temperature is 

w _ s r s i tz. 1 "I 

(H. P. Cap. cold) 81 

(81) 



The low-pressure capacity stated in terms of high-pressure capacity hot, 

is actually discharged is 

i_ 

(L. P. Cap.) = (H. P. Cap. hot)# p2 s R P i, ..... (82) 
whence 

- 
per cubic foot hot gas discharged 

(H. P. (Tap. 



-2 . (83) 



WORK OF COMPRESSORS 



105 



Examination of Fig. 28 will show without analysis that there must be some 
bit-receiver pressure at which least work will be required. For if the receiver 
p ; ssure approached PD then the compression would approach single stage and 



II 







looj saunbs aod spunod uj saansseaa: 



|je compression line approach BCG. The same would be true as the receiver 
jessure approached P g = P e , whereas at any intermediate point C, intercooling 
tuses the process 'to follow BCDE with a saving of work over single-stage 
oeration represented by the area DCGE. This area being zero when C is at 



106 ENGINEERING THERMODYNAMICS 

either B or G, it must have a maximum value somewhere between, and 
pressure at which this least-compressor work will be attained is the bcst-recei 
pressure. 

By definition the best-receiver pressure is that for which W is a minimu 
or that corresponding to 

dp," 

Performing this differentiation upon Eq. (78), equating the result to ze 
and solving for P c , 



(Best rec.pr.) = (P b P e )*~ [(sup. pr.) (del. pr)]* (*; 

Substituting this value in the general expression for work Eq. (78), noting tl 






Eq. (85) is the general expression for two-stage work with perfect int- 
cpoling at best-receiver pressure in terms of pressures and volumes. Sn- 
stituting the symbols for the pressures and volumes and noting that aso 
Cycle 1, 

F & =(L. P. Cap.) and F e =(H. P. Cap. hot) and using; -(#) for (~\ 



= 288--(sup.pr.)(L. P. Cap.) 



This equation gives the same value as Eq. (85), but in terms of different u 
It should be noted here that the substitution of best-receiver pressure in 

expressions for the two stages preceding Eq. (78), will show that the work d 

in the two cylinders is equal. 

The work per cubic foot of low-pressure gas, from Eq. (86) is, 



(L.pcap.) 



To transform Eq. (85) into a form involving delivery volumes, use the r 
tion from the diagram, 



i 

pj\p.. 



WORK OF COMPRESSORS 107 

Whence 



e \p b j\p d , 

which for the best-receiver pressure becomes 

8+1 

Substituting in Eq. (85), 

Introducing the symbols, 

. P. Cap. hot)E p ~2s~ R p ~lte~ 1 , . . (89) 



and 

s-1 



- ' ' (90) 



The volume of gas discharged at the higher pressure when reduced to its 
original temperature will become such that 

(L.P.Cap.) = P e 
(H. P. Cap. cold) P b py 

or 

(sup.pr.) (L. P. Cap.) = (del.pr.) (H. P. Cap. cold), . . . (91) 

which may be substituted in Eq. (86), 



.pr.)(H.P. Cap. cold)/2r-l . . . (92) 
from wrrich the work per cubic foot of gas delivered and cooled is, 

w 



(H.P. Cap. 



108 ENGINEERING THEEMODYNAMICS 

Example 1. Method of calculating diagram, Fig. 28. 
Assumed data: 

Fa =0 cu.ft. P a =2116 Ibs. per square foot. 



F/=0 cu.ft. P c =P d = VP a P e =6172 Ibs. sq.ft. 
F 6 = 5 cu.ft. P/ = P.=P,= 18,000 Ibs. sq.ft. 



5 = 1.4. 

To obtain point C: 



or 

/. F c = 2.36 cu.ft. P c = 6172 Ibs. sq.ft 

To obtain point D: 



.*. . F d = 1.71 cu.ft. Po- = 6172 sq.ft. 
To obtain point E: 



but by definition 



hence, 

Fe = 1.71 -2.14 = .8 cu.ft. P e = 18000 Ibs. sq.ft. 

Example 2. To compress 5 cu.ft. of air from one atmosphere (2116 Ibs. per sq\ 
foot) to 8.5 atmospheres (18,050 Ibs. per square foot) in two stages with best-receive 
pressure and perfect intercooling requires how much work? 



W =288--(sup.pr.)(L. P. Cap.)(fl p -1), 
s 1 

(sup.pr.) =14.7. (L. P. Cap.) =5. fl^-8.5. 

/ irJL \ 
.'. TF=288X3.463X14.7X5X (8.5 2a -ij =26,800 ft.-lbs. 






WORK OF COMPRESSORS 109 



Prob. 1. Air at 14 Ibs. per square inch absolute is compressed to 150 Ibs. per square 
nch absolute by a two-stage compressor. What will be the work per cubic foot of air 
elivered? What will be the work per cubic foot if the air be allowed to cool to the 
riginal temperature, and how will this compare with the work per cubic foot of sup- 
ilied air? Best receiver-pressure and perfect intercooling are assumed for the above 
ompressor, s = 1.4. 

Prob. 2. A compressor receives air at atmosphere and compresses it to half its vol- 
.me, whereupon the air is discharged to the cooler and its temperature reduced to the 
riginal point. It then enters a second cylinder and is compressed to 80 Ibs. absolute. 
Vliat will be the work per cubic foot of supplied air in each cylinder and how will 
he work of compressing a cubic foot to the delivery pressure compare with the work 
.one if compression were single stage, compression being adiabatic. 

Prob. 3. Air is to be compressed from 15 Ibs. per square inch absolute to 10 times 
jhis pressure. What would be the best-receiver pressure for a two-stage compressor? 
v flow many more cubic feet may be compressed per minute in two stage than one stage 
i|y the same horse-power? 

Prob. 4. A manufacturer sells a compressor to run at best-receiver pressure 
tfhen (sup.pr.) is 14 Ibs. per square inch absolute and (del.pr.) 100 Ibs. per square inch 

1 -.bsolute. What will be the work per cubic foot of supply-pressure air done in each cylin- 
j,-ier? Another compressor is so designed that the receiver pressure for same supply 
pressure and delivery pressure is 30 Ibs. per square inch absolute, while a third is so 

lesigned that receiver pressure is 50 Ibs. per square inch absolute. How will the 
rork done in each cylinder of these machines compare with that of first machine? 

Prob. 5. For an ice machine a compressor works between 50 and 150 Ibs. per square 
ach absolute. It is single stage. Would the saving by making compression two stage 
t best-receiver pressure amount to a small or large per cent of the work in case of single 
tage, how much? 

Prob. 6. A compressor has been designed to compress 1000 cu.ft. of carbon dioxide 
ter minute from 15 to 150 Ibs. per square inch absolute. What horse-power will be 
;equired at best-receiver pressure? Should delivery pressure change to 200 Ibs., what 
ower would be required? To 100 Ibs. what power? 

Prob. 7. A gas-compressing company operates a compressor which has to draw 

00 2 gas from a spring and compress it to 150 Ibs. per square inch gage. In the morning 
ressure on the spring is 10 Ibs. gage, while by evening it has dropped to 5 Ibs. absolute. 

pf the compressor was designed for the first condition, how will the high-pressure 
Capacity cold and horse-power per cubic foot of high-pressure gas at night compare 
S-irith corresponding values in morning? Assume a barometric reading. 

Prob. 8. On a mining operation a compressor is supplying a number of drills and 
moists with air at 150 Ibs. per square inch absolute, the supply pressure being 14 Ibs. 
ijjVhat will be the difference in horse-power per cubic foot of delivered air at compressor 

jud per cubic foot received at drills if air is a long time in reaching drills? 

Prob. 9. With a best-receiver pressure of 40 Ibs. per square inch absolute and a 

upply pressure of 14 Ibs. per square inch absolute, what horse-power will be required 
1 o compress and deliver 1000 cu.ft. of high-pressure air per minute at the delivery 

>ressure for which compressor is designed and what is that delivery pressure? 

10. Two-stage Compressor, with Clearance, Perfect Intercooling Expo- 
icntial Compression, Best-receiver Pressure, Equality of Stages, (Cycle 6). 
Work and Capacity in Terms of Pressures and Volumes. The two-stage expo- 



110 



ENGINEERING THERMODYNAMICS 



nential compressor with clearance and perfect intercooling is represented by 
the PV diagrams Figs. 29, 30, 31, which are clearly made up of two single-stage 
compression processes, each with clearance. 






^ooj 94Bnbg aad -sqi ui seanssaaj; w 



I 



a 



Applying Eq. (57) to the two stages and supplying proper subscripts; 
referring to Fig. 29, 



8-1 

-F a )[^ ' -l]. . 



(first stage) 



WORK OF COMPRESSORS 111 

*z_ J 

1 . . (second stage) 


[f the condition of" perfect intercooling be imposed, it is plain that since 

n weight of gas entering the second stage must equal that entering the first 
jj;e, and the temperature in each case is the same, 



jj noting also that 



(94) 



B (94) is the general expression for the work of two-stage exponential 
:cipressor with perfect intercooling, P c being the receiver pressure. 

p 

As in Section 9, let (R P i) be the pressure ratio - for the first stage and(72?2) 

"b 

Pe 

ih pressure ratio for the second stage and using instead of P& its equivalent 

PC 

A (sup.pr.) Ibs. per square inch. 

1 1 . . (95) 



fjch is identical with (79), showing that for two-stage compressors with perfect 
nrcooling (as for single stage, Section 7), the work for a given low-pressure 
aicity is independent of clearance. 

The work per cubic foot of gas supplied is given by Eq. (80); per cubic 
0) of cold gas delivered by Eq. (81) and per cubic foot hot gas delivered by 
I (83). 

The reasoning regarding best-receiver pressure followed out in Section 9, will 

dW 

1 again in this case, and by putting = in Eq. (94), and solving for P c 

dP c 

all again be found that best-receiver pressure will be 

(best-rec. pr.) = (P 6 P e )* ......... (96) 

Substitution of this value for P c in Eq. (94), gives the following expression 
Dvork of the two-stage exponential compressor with best-receiver pressure, 



(97) 



112 



ENGINEERING THERMODYNAMICS 






which may be expressed in terms of supply pressure, pounds per square in 
low-pressure capacity, cubic feet, and ratio of compression, 




^ (sup.pr.) (L. P. Cap.) \R P '2. * - 1 J , 



which is the same as Eq. (86). 

Substitution of the value of best -receiver pressure in the expression for 



WORK OF COMPRESSORS 11^ 

Ork-of the two stages separately will show the equality of work done in the 
ispective stages for this case with clearance. 



r ork per cubic foot gas supplied to compressor is 

w 






(gg) 



ork per cubic foot of high-pressure gas hot is 



W s lir irJ 1 

288 ^!^^*' 2s I* " 



The work per cubic foot of air delivered and cooled to its original tem- 
iture is, 

w 9 r ~ i i 

(H. P. Cap. cold) - ^-^(deLpr.) |^~ 1 J , - . (101) 

Due to the fact that clearance has no effect upon the work per cubic foot 
substance, as previously noted, Eqs. (99), (100) and (101) are identical with 
*7), (90) and (93). 

11. Two-Stage Compressor, any Receiver Pressure, Exponential Compres- 

lion. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure and 

Eorse-power, in Terms of Dimensions of Cylinders and Clearances. Referring to 

frig. 29, let DI be the displacement of the first stage cylinder in cubic feet = 

Vb Vt), D2 the displacement of the cylinder of the second stage in cubic 

lpet = (Fd-- Vf)j ci the clearance of the first stage, stated as a fraction of the 

Displacement of that cylinder, so that the clearance of the first stage cubic 

9et = ciDi, and that of the second stage = c^Di. 

The low-pressure capacity of the first stage (L. P. Cap.) in cubic feet is 
VbVa), and, as for the single-stage compressor, is expressed in terms of dis- 
placement, clearance and ratio of compression of the first stage as follows, see 
"Sq. (64); 



(L.P. Cap.i)=Dil+ci-ci# P i =DiE vl ...... (102) 

For the second stage, the low-pressure capacity (L. P. Cap. 2) is 



jjnd is equal to 



I 
(L.P. Czp.2)=D 2 l + c 2 -C2Rp2 s =D 2 E C2 .... (103) 



Volumetric efficiency of the first stage is given by 

i 

* ....... (104) 



t I 

I 
metric efficiency for second stage 



ENGINEERING THERMODYNAMICS 





(105 




I 

6 
I 

02 

6 



2 


It may be required to find the receiver pressure (incidental to the finding c ; 
work or horse-power) for a compressor with given cylinder sizes and deliver! 
pressure. The condition assumed of perfect intercooling stipulates that 



(L. P. Cap.i)(sup.pr.) = (L. P. Cap. 2 ) (rec.pr.), 



WORK OF COMPRESSORS 

'hence 

.,(L.P.Cap.i) 



(106) 



If the volumetric efficiencies are known or can be sufficiently well approx- 
.iiated this can be solved directly. If, however, E v \ and E& are not known, 
ut the clearances are known, since these are both dependent upon the receiver 
ressure sought, the substitution of the values of these two quantities will give 



(rec.pr.) = (sup.pr.) - ^-, . . . (107) 

D2\ l~f"C2 C2\ ' 

\ rec.pr. 

,n expression which contains the receiver pressure on both sides of the equation. 
: f his can be rearranged with respect to (rec. pr.), but results in a very complex 
impression which is difficult to solve and not of sufficient value ordinarily to 

varrant the expenditure of much labor in the solution. Therefore, the relations 

ire left in the form (107). It may be solved by a series of approximations, 

,he first of which is 

(rec.pr.) = (sup.pr.)^ approx. ...... (108) 

L/2 

Vith this value for the receiver pressure, substitution may be made in the sec- 
ond member of the Eq. (107), giving a result which will be very nearly correct, 
if desirable, a third approximation could be made. 

To find the work of a two-stage exponential compressor in terms of displace- 
ment of cylinders, supply pressure, receiver pressure and delivery pressure, 
>ounds per square inch, and volumetric efficiency of the first stage, E v i, from (79) 
r (94) , 



. . . ,(109) 

a J. L J 

in which 

(rec.pr.) (del.pr.) 



v -f ~A p \. 

(sup.pr.) (rec.pr.) 

,Fo solve this the receiver pressure must be found as previously explained and 
;he volumetric efficiency must be computed by Eq. (104) or otherwise be known. 
It is impracticable to state work for this general case in terms of displace- 
ment and clearances directly, due to the difficulty of solving for (rec. pr.) in Eq. 



116 ENGINEERING THERMODYNAMICS 






(107). It may, however, be stated purely in terms of supply and delivery pres 
ejures, in pounds per square inch, displacement, in cubic feet, and volumetri 
efficiencies, as follows: 
From Eq. (106), 



and 

_del.pr. ..D 2 E v2 _ D 2 E v2 
Kp2 ~ sup.pr. X Z>i#,i ~ ^DiEn' 

Hence 



The mean effective pressure of the two-stage compressor referred to the lowl 
pressure cylinder is found by dividing the work of the entire cycle Eq. (110), b 
the displacement of the first-stage cylinder, and by 144, to give pounds pe 
square inch. 

m.e.p. referred to first-stage cylinder, pounds per square inch is, 



It is well to note that this may also be found by multiplying (work don: 
per cubic foot of gas supplied) by (volumetric efficiency of the first stage, E v \] 
and dividing the product by 144. 

In terms of the same quantities, an expression for indicated hbrse-powe 
may be given as follows: 

-! a-1 



s- (sup.pr.) | p v . 

f f, ~ 2 , 



1 
~ 2J , 



where n is the number of cycles completed per minute by the compressor 
For n may be substituted the number of revolutions per minute, divided by th 
revolutions per cycle, 

N 



The horse-power per cubic foot of gas supplied per minute is 

'I.H.P s (sup.pr.) \(DiE vl \-s- / D2E v2 \-*- J 

n(L.P.Cap.) "7-^1 ^29T" [\D 2 E v2 ) f ("'D^J 



WORK OF COMPRESSORS 117 

; orse-power per cubic foot of gas delivered and cooled per minute. 

s -i .-i 



n(H. P. Cap. cold) s-l 229.2 \D 2 E 
orse-power per cubic foot of hot gas delivered per minute 

- LH.P. s sup.pT.(D l E 9 i\ L ir L 

i n(H.P.Cap. hot) s-l 229.2 \D 2 E, 2 ) 



R _ 

" 



7^_ 

2 - 



lor the case where clearance is zero or negligible, these expressions may be 
mplified by putting E,? and E i equal to unity. 



_ 

s (sup.pr.) [/DA . ,( R D 2 \ _J 

~i=i"M95~ ;I LW f v w 

,H.P. per cubic foot, gas supplied per minute 



D\ ] 

J' 



n(L. P. Cap.) ~ 
H.P. per cubic foot gas delivered and cooled per minute 

I.H.P. (del.prOr/gA-?' /Da\-'- 



n(H. P. Cap. cold)" s-l 229.2 
.P. per cubic foot hot gas delivered per minute 



Example 1. Method of calculating diagram, Figs. 29, 30, 31. 
\(.Sf>umed data. 

p a =p b =2116 Ibs. per square foot; 
p c = p d =p h =p k =:Ql72 Ibs. per square foot. 
P g =P e =P f = 18,000 Ibs. per square foot. 
7(H. P.) = 7.5 per cent; C7(L. P.) 7.5 per cent; s = 1.4; L. P. Capacity =5 cu.ft, 



118 ENGINEERING THERMODYNAMICS 

To obtain point A". 

From formula Eq. (64), 

i 

5 = Di(l + .075-.075X2.14), hence Di =5.45 cu.ft. 

Cli =V k = 5.45 X .075 = .41 cu.ft. 
/. Ft = .4 cu.ft.; P t =6172 Ibs. sq.ft. 

To obtain point A : 

i 

F a = F*(^Y' 4 = .4X2.14 = .856 cu.ft. 
\* </ 

.'. Fa =.85 cu.ft.; P a =2116 Ibs. sq.ft. 
To obtain point B: 

F 6 = F +5 = .85 +5 =5.85 cu.ft.; P 6 =2116. Ibs. sq.ft. 

To obtain point C: 



' =5.85 -s-2.14 =2.73, 

4 

F c =2.73 cu.ft.; P c = 6172 Ibs. sq.ft. 



To obtain point D: 

Volume at D is the displacement plus clearance of H. P. cylinder. This cannot I 
found until the capacity is known. The capacity is the amount gas which must b 
taken in each stroke and which is also the amount actually delivered by L. P. cylindc 
cooled to original temperature. The amount of cool gas taken in by the second cylind 
is 



L V 



(L. P. Cap. 2 ) = ~(L- P. Cap.O = X5 = 1.7 cu.ft. 

But 



cu.ft. 

. 14=2.02 cu.ft. 
P d = 6172 Ibs. sq.ft. 
Other points are easily determined by relations too obvious to warrant setting dow: 

Example 2. What will be the capacity, volumetric efficiency and horse-power p 
1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for the follow j 
ing compressor: Two-stage, double-acting cylinders, 22| and 34jX24in., runnirf 
at 100 R.P.M. High-pressure clearance 6 per cent, low-pressure 4 per cent. Supp i 



WORK OF COMPRESSORS 119 

essure 14 Ibs. per square inch absolute. Delivery pressure 115 Ibs. per square inch 
>solute. 
The capacity will be the cylinder displacement times the volumetric efficiency. 

Pi = displacement of a 34ix24 in. cylinder, or 12.8 cu.ft. and Z>2 = displacement of 
22|+24 /r cylinder or 5.4 cu.ft. To obtain the volumetric efficiency, make use of 
;>proximation of formula Eq. (108), 

D 12 8 

(rec.pr.) = (sup.pr.) =14 X~-r =33.2 Ibs. sq.in, 
L/z o.4 

id then by Eq. (107) checking, 



(rec.pr.) =14X i^ ^ ' =35.1 Ibs. sq.in. 



__ 

E n = 1 +ci -Ci(R pl ) s from Eq. (104), 

i 
= l+.04-.04x(2.5)~ s =96.8 per cent 

herefore the capacity will be, 

200 X 12.8 X. 968 =2480 cu.ft. per minute; 
E C2 = 1 +c 2 -C 2 (# P2 ) from Eq. (105), 
= 1+.06-.06X(3.28)' 714 =92 percent. 

from Eq. (113), I H.P. per cu.ft. (sup.pr.) air per minute is, 

s-l s-l 



s sup.pr. [ (DJti\ ( K D 2 E V2 \ . ] 
= r-l "229T L \D 2 E~J f V*KR/, 

1.4 14 r/12.8x.968\-286 / ^4x^\.286 n 
T X 229^2 L \^4XlJ2"/ 22 12.8X.968/ 



/"hence horse-power- per 1000 cu.ft. of free air per minute is, =150. 
From Eq. (115) horse-power per cubic foot (del.pr.) air, hot = that of (sup.pr.) 

( L - j or 5.85 times that of (sup.pr.) air. 

,'. Horse-power per 1000 cu.ft. of hot (del.pr. air) = 150 X5.85 =877. 

i Prob. 1. A two-stage double-acting compressor has volumetric efficiencies as shown 
y cards of 98 per cent and 90 per cent for the high- and low-pressure cylinders respect- 
/ely. It is running at 80 R.P.M. and compressing from atmosphere to 80 Ibs. per square 
ich gage. If the cylinders are 15jx25i Xl8 ins., and speed is 120 R.P.M., what 



120 ENGINEERING THERMODYNAMICS 

horse-power is being used and how many cubic feet of free and compressed air (hoi 
and cold) are being delivered per minute, when s equals 1.41? 

Prob. 2. What horse-power will be needed to drive a two-stage compressor 10 \ ins 
and 16i Xl2 ins., double acting, with 5 per cent clearance in each cylinder at 160 R.P.M 
when the supply pressure is atmosphere, delivery pressure 100 Ibs. per square incl 
gage, when s 'equals 1.35? 

Prob. 3. A thousand cubic feet of free air per minute must be compressed by a two- 
stage compressor to 80 Ibs. per square inch gage from a supply pressure of 10 Ibs. pei 
square inch absolute. The volumetric efficiencies for the high- and low-pressure cylin- 
ders are 85 per cent and 95 per cent respectively, and the receiver pressure is 25 Ibs 
per square inch absolute. What will be the displacement of each cylinder and th< 
horse-power per cubic foot of (sup.pr.) air? 

Prob. 4. How many cubic feet of free air can be compressed in two-stage compres 
sor 18ix30i X24 ins. with 5 per cent clearance in high-pressure cylinder and 3 per cem 
in low if (sup.pr.) is atmosphere and (del.pr.) 80 Ibs. per square inch gage? How woulc 
the answer be affected if clearance were taken as zero? Take s = 1.41. 

Prob. 5. The volumetric efficiency of the low-pressure cylinder is known to be 95 pe, 
cent, and of the high-pressure cylinder 85 per cent. The cylinder sizes are 15| X25^ xli 
ins. and speed is 120 R.P.M. What horse-power must be supplied to the machine ij 
the mechanical efficiency is 80 per cent and the pressure ratio 10 with a (sup.pr.) of on 
atmosphere? 

Prob. 6. A compressor runs at 120 R.P.M. and is double acting. It is compressing 
air from 14 Ibs. per square inch absolute to pressures ranging from 70 Ibs. per squar< 
inch gage to 100 Ibs. per square inch gage. The cylinders are 20}x32ix24 ins., am 
clearances 8 per cent and 4 per cent. Find the approximate receiver pressure, capacity 
and horse-power for the range of discharge pressure, for s = 1.3. 

Prob. 7. The volumetric efficiency of the low-pressure cylinder of a two-stage com 
pressor is known to be 95 per cent, the receiver pressure as shown by gage is 40 Ibs. 
delivery pressure 100 Ibs., and supply pressure one atmosphere. What will be th> 
horse-power if the machine runs at 120 R.P.M. and the low-pressure cylinder is 18 X 12 in. 
s = 1.4. 

Prob. 8. An air compressor appears to require more power to run it than shouL 
be necessary. It is a double-acting 18x30x24 in. machine running at 100 R.P.M 
The volumetric efficiencies are 85 per cent and 90 per cent respectively and supply an< 
delivery pressures 14 Ibs. and 110 Ibs. per square inch, both absolute. What would b f 
the minimum work per cubic foot of (sup.pr.) air, per cubic foot of (del.pr.) air, hot an</. 
cold, for adiabatic compression? 

Prob. 9. The efficiency of the driving gear on an electric-driven compressor i^ 
75 per cent. Power is being supplied at the rate of 150 H.P. How much air shoul*; 
be compressed per minute from 4 Ibs. per square inch absolute to 100 Ibs. per squarl 
inch gage, if the receiver pressure is 35 Ibs. per square inch absolute and the low-pressur i 
volumetric efficiency is 90 per cent, s being 1.4? 

12. Two-stage Compressor with Best Receiver Pressure Exponents 
Compression. Capacity, Volumetric Efficiency, Work, Mean Effective Pres 
sures and Horse-power in Terms of Dimensions of Cylinders and Clearances 
For the two-stage exponential compressor with or without clearance, and pei 
feet intercooling, the best-receiver pressure was found to be (Eq. 84), 

(best-rec.pr.) = [(sup.pr.) (del.pr.)]* (12C 



WORK OF COMPRESSORS 121 

This expression Eq. (120) for best-receiver pressure makes it possible to 
aluate R P i and R p2 as follows: 



sup.pr. sup.pr. L su P-P r 



(del.pr.) (del.pr.) 

for (best-rec. pr.) = 77 >. = r- f w , , . u 

(best-rec.pr.) [(sup.pr.) (del.pr.)]' 



r/deLprAl t = 
L \sup.pr.y J 



The use of these values for R p \ and R P 2 in the expressions previously given 
}>r volumetric efficiency for the general case, Eqs. (104) and (105) results in 

iolumetric efficiency, first stage 

_i 

ad volumetric efficiency, second stage 

i_ 
i 

The work was found to be represented by Eq. (98), which may be stated 
terms of displacement and volumetric efficiency of the first stage, as follows: 



.... (125) 

here R p = ( G ' pr \ and where (sup.pr.) is in pounds per square inch, 
(sup.pr.; 

If the clearance is known for the first stage this becomes by the use of 
q. (104), 



-2r-l, .-. . (126) 
s 

Kich is a direct statement of the work of a two-stage adiabatic compressor 
ith perfect intercooling in terms of supply pressure and delivery pressure, 
ounds per square inch, displacement, cubic feet and clearance as a fraction 



122 ENGINEERING THERMODYNAMICS 

of displacement, provided the cylinder sizes and clearances are known to be sud 
as to give best-receiver pressures. 

The mean effective pressure reduced to first-stage displacement, in pounds 
per square inch, may be derived from either Eq. (125) or (126) by dividing tl 
work by the displacement of the first-stage cylinder, and again dividing by 



2s / _L\ r L~ 1 

= -^ (sup.pr.)l l+ci-ci/^2* }\R P 2 S -1 



(127. 



Since the work done is equally divided between the two cylinders when best 
receiver pressure is maintained, the mean effective pressure, in pounds pei 
square foot, for each cylinder will be, one-half the total work divided by th( 
displacement of the cylinder in question, 

w s r s ~ i i 

m.e.p., first stage = 2gg^= -j(sup.pr.)^i| /^ 2. -l|. ..... (128 



Note that this is one-half as great as the m.e.p. of the compressor reducec 
to first stage, (127), 



w s , N Di r ^ 

m.e.p., second stage = 2ggyT- = ^2Y(sup.pr.)^- 7 P iLR 1? 2s 1 1 . . . (129 

But 

D\E c i [ 1 

(sup.pr.)^ ^ = (rec.pr.) = (sup.pr.)(del.pr.) p, 

Lf2^v2 

whence, 

s If 1 f 1 

.e.p., second stage= (sup.pr.)(del.pr.) U^LRp 2 1. . . . (II 



m 



It is next necessary to investigate what conditions must be fulfilled to obtai 
the best-receiver pressure, the value of which is stated, Eq. (120). The conditiol 
of perfect intercooling provides that the temperature of the gas entering tr 
second stage is the same as that entering the first stage, and hence that tl ! 
product (volume entering second stage) X (pressure when entering second stage ' 
must be equal to the product (volume entering first stage) X (pressure of suppl 
to first stage), or 

(L. P. Cap. 2 )(rea pr.) = (L. P. Cap. i) (sup. pr.), . . . (12 



WOEK OF COMPEESSOES 123 

Ombining with Eq. (120) 

(L. P. Cap.Q [(sup.pr.) (del.pr.)]* [ (del.pr.) ]*_ . 
(L. P. Cap. 2 ) ~ (sup.pr.) [ (sup.pr.) J " 

(1) (2) r (3) J-, 

^_(L.P.C*p. 1 )_D l E vl _ Dl [ 1 + Cl - ClR ^\ 

r -i * * * * 

: 1 4~ C2 C2^p2s I 




(L. P. Cap. 2 ) -*-,* D 



un this three-part equation proper values may be found to fulfill require- 
.ents of best-receiver pressure for : 

1. The ratio of capacities for a given ratio of pressures, or conversely, the 

ratio of pressures when capacities are known; 
! 2. The ratio of cylinder displacements for known volumetric efficiencies; 

3. The ratio of cylinder displacements when the clearances and ratio of com- 

ression are known, or conversely, with known displacements and clearances 

jiie ratio of pressures which will cause best-receiver pressure to exist. This 

ist case in general is subject to solution most easily by a series of approxi- 

lations. 

There is, however, a special case which is more or less likely to occur in prac- 
ice, and which lends itself to solution, that of equal clearance percentages. If 
i = C2 the parenthesis in the numerator of part (3) of Eq. (132) becomes equal 
D the parenthesis in the denominator, and evidently the volumetric efficiency 
if the two cylinders are equal, hence for equal clearance percentages in the two 
ages, 



- ...... 

f case which leads to the same expression, Eq. (133), is that of zero clearance, 
I condition that is often assumed in machines where the clearance is quite 



LThe work per cycle, Eq. (126), when multiplied by the number of cycles 
rformed per minute, n, and divided by 33,000, gives 

-l), (134) 



from which are obtained the following: 
[.H.P. per cubic foot supplied per minute 



n(L.P. Cap.) 8-1 114.6 

f .H.P. per cubic foot delivered and cooled per minute 

I.H.P. s (del.pr.) 

n(H. P. Cap. cold) s-1 114.6 ( 






124 [ENGINEERING THERMODYNAMICS 

and I.H.P. per cubic foot delivered hot per minute 



I.H.P. . (sup.pr.) +! =! 

2s (K p 2s 1). . (137. 



n(H. P. Cap. hot) s- 1 114.6 

These expressions, Eqs. (165), (166) and (167) are all independent of clear 
ance. 

Example. What will be the capacity, volumetric efficiency and horse-power pe 
1000 cu.ft. of free air and per 1000 cu.ft. of hot compressed air per minute for th 
following compressor for s = 1.4? Two-stage, double-acting, cylinders 22| X34^ X24 ins. 
running at 100 R.P.M. Low-pressure clearance 5 per cent, high-pressure clearanc 
such as to give best-receiver pressure. Supply pressure 15 Ibs. per square inch abso 
lute, delivery pressure 105 Ibs. per square inch absolute. 

Capacity will be cylinder displacement times low pressure volumetric efficiency, ori 
200Z>iX#,i. 

Di = 17.5 cu.ft. 

En from Eq. (123) - (1 +Ci -dR P 2~s) 

= 1+.05-.05X7' 357 =95 per cent. 

Therefore low pressure capacity = 200 X 12.8 X. 95 =2430 cu.ft. per minute. 
Horse-power per cubic foot of (sup.pr.) air per minute is from Eq. (135) 

.<? siirt rr * 1 



1 



Therefore, horse-power per 1000 cu.ft. of sup.pr. air = 160. 

Horse-power per cubic foot of (del.pr.) air, hot, is from Eq. (137) 

1+8 

R p 2s times power per cu.ft. of (sup.pr.) air, 

hence, 

160 X5.3 =850 =horse-power per 1000 cu.ft. of (del.pr.) air, hot, per minute. 

Problem Note. In the following problems, cylinders are assumed to be proportione 
with reference to pressures so as to give best-receiver pressure. Where data conflic 
the conflict must be found and eliminated. 

Prob. 1. Air is compressed adiabatically from 14 Ibs. per square inch absolute t 
80 Ibs. per square inch gage, in a 20j X32| X24 in. compressor, running at 100 R.P.M 
the low-pressure cylinder has 3 per cent clearance. What will be horse-power J:< 
quired, to run compres or and what will be the capacity in cubic feet of low prcssu: 
and in cubic feet of (del.pr.) air? 

Prob. 2. What must be the cjdinder displacement of a two-stage compressor with 
per cent clearance in each cylinder to compress 500 cu.ft. of free air per minute fro) 
14 Ibs. per square inch absolute to 85 Ibs. per square inch gage, so that s equals I/, 
What will be the horse-power per cubic foot of (del.pr.) air hot and cold? 

Prob. 3. A two-stage compressor is compressing gas with a value of s = l. 
from 10 Ibs. per square inch gage to 100 Ibs. per square inch gage. The cylinders ai 
18^X301x24 ins., and speed is 100 R.P.M. If the clearance is 5 per cent in the lov 



WOEK OF COMPRESSORS 125 

I assure cylinder and 10 per cent in the high, what will be the cubic foot of (sup.pr.) gas 
i ndled per minute and what will be the horse-power at best receiver pressure? 

Prob. 4. A manufacturer states that his 20 jX 32^x24 in. double-acting compres- 
gr when running at 100 R.P.M. at sea level will have a capacity of 2390 cu.ft. of free air 
jr minute, pressure range being from atmosphere to 80 Ibs. per square inch gage. At 
tst-receiver pressure what clearance must the compressor have, compression being 
i iabatic? 

Prob. 5. The cylinder sizes of a two-stage compressor are given as 10ixl6|xl2 
13., and clearance in each is 5 per cent. What will be the best-receiver pressures when 
(i crating between atmosphere and following discharge pressures, 60, 70, 80, 90, 100 
td 110 Ibs. per square inch gage, for s equal 1.4? 

Prob. 6. 1500 cu.ft. of air at 150 Ibs. per square inch gage pressure are needed per 
inute for drills, hoists, etc. The air is supplied from 3 compressors of the same size 
(d speed, 120 R.P.M. Each has 4 per cent clearance in each cylinder. What will 
! sizes of cylinders and the horse-power of the plant for best-receiver pressure, when 
JM.41? 

Prob. 7. The cards taken from a compressor show volumetric efficiencies of 95 per 
mt and 80 per cent .in low- and high-pressure cylinder respectively. What will be 
iel.pr.) for best-receiver pressure if compressor is 15lx25jXl8 ins., and (sup.pr.) 15 
jk per square inch absolute to 10 Ibs. absolute, and what will be the work in each case, 
being 1.35? 

Prob. 8. A manufacturer gives a range of working pressure of his lOjx 16^X12 in. 
<mpressor from 80-100 Ibs. per square inch p-age. If clearances are, low 4 per cent, 
Ugh 8 per cent, and (sup.pr.) is atmosphere, find by trial which end of the range comes 
Barest to giving best-receiver pressure? If clearances were equal which would give 
list-receiver pressure? 

Prob. 9 A 16 J X25j Xl6 in. compressor is rated at 1205 cu.ft. free air per minute at 
;i5 R.P.M. at sea level. What would be the clearance if compressor were compressing 
a* from atmosphere to 100 Ibs. gage at sea level? With same clearance what would be 
ie size of a low-pressure cylinder to give the same capacity at altitude of 10,000 ft. with 
e same clearance and the same (del.pr.) , best-receiver pressure always being maintained? 

13. Three-Stage Compressor, no Clearance, Perfect Interceding Expo- 

sntial Compression (Cycle 7), Best Two Receiver Pressures, Equality of 

;tages. Work and Capacity, in Terms of Pressures and Volumes. The three- 

age exponential compressor cycle with no clearance, perfect intercooling Cycle 

il is shown in Fig. 32. The net work area, ABCDEFGHJKA , is made up of 

Iliree areas which may be computed individually by the formulae for single stage 

i?q. (48), provided the requisite pressures and volumes are known, as follows: 

1 

W = ~ri P V [(j?) * - 1 ] (first stage) 

s-l 



~Tl P V [(j?) S ~ l ] (second stage) 



s l 



(third stage) 



. . (138) 



126 ENGINEERING THERMODYNAMICS 

But the condition of perfect intercooling provides that for no clearance, 



Q- 




&& 



-(JdJ-oaa) 



m 



\q; aa^nbg jsd 'sc 

and it may be noted that Pd = P C) and P f =P e . Accordingly, 

s-l s-1 f s-1 



Pressures in this expression are in pounds per square foot. 



. . (14 



WORK OF COMPRESSORS 127 

Changing the equation to read in terms of supply pressure pounds per square 
ich, low-pressure capacity cubic feet, and ratios of pressures, first stage 
, second stage (R P 2) and third stage (R P z), it becomes 



/ork done by three-stage compressor, perfect intercooling 

r s-i s-i 9-1 

.P. Cap.) \(R p i) ^~ + (R v 2)~* +(Rpz) -31, (141 



this the following expressions are derived: 
rk per cubic foot supplied 

si s-l * i 

--3]. . . (142) 



(143) 



rk per cubic foot gas delivered and cooled 






1 -3] . 



ork per cubic foot gas, as delivered hot 

s-l 



(144) 



Best Two Receiver Pressures. Referring to Fig. 32, P c is the pressure in the 

fj-st receiver (1 rec.pr.) and P e is the pressure in the second receiver, (2 rec.pr.). 

3 is evident that if either receiver pressure be fixed and the other is varied, 

ie work necessary to compress a given initial volume of gas will be varied, 

j&d will have a minimum value for some particular value of the varying receiver 

f {essure. By a variation of both receiver pressures a minumim may be found 

jlr the work when both receiver pressures have some specific relation to supply 

:b.d delivery pressures. For instance, assume that P c is fixed. Then a change 

i P e can change only the work of the second and third stages, and the three- 

eige compressor may be regarded as consisting of 

One single-stage compressor, compressing form P& to P c . 
One two-stage compressor, compressing from P c to P . 

. In this two-stage compressor, best-receiver pressure is to exist, accord 
i. to Eq. (84), 

P e = (best 2 rec .pr.) = (P C P,)*. . . . (145) 



128 ENGINEERING THERMODYNAMICS 

Similar reasoning, assuming P e fixed and making P c variable, would shoi 
that 

P c =(best 1 rec.pr.) = (P e P 6 )*. ...... (146 

Eliminate P c from Eq. (145) and the expression becomes, 

P e = (best 2 rec.pr.) = (P,P 2 ) = [(sup.pr.Xdel.pr.) 2 ] * (1 

Similarly, from Eq. (146) 

P c = (best 1 rec.pr.) = (P, 2 P g ) = [(sup.pr.) 2 (del.pr.)] * (14* 

From these expressions may be obtained, 



p. p p 

z& r c r e 

or 

Rpi =Rp2 R 



Substitution in Eq. (140) gives, 

Work, three-stage, best-receiver pressure no clearance 



Arranging this equation to read in terms of supply pressure, pounds per squa . 
inch, low-pressure capacity, cubic feet, and ratio of pressures 

Work, three-stage best-receiver pressure 



sl 



= 432-^-(sup.pr.)(L.P. Cap.)(#/37 -1), . . . (u 
s i 



The work of the compressor is equally divided between the three staty 
when best-receiver pressures are maintained, which may be proven by substi 
tion of Eq. (149) in the three parts of Eq. (138), and 

Work of any one stage of three-stage compressor with best-receiver pressure 

s-l 



WORK OF COMPRESSORS 129 

From Eq.(151), may be derived the expressions for work per unit of capacity. 
\Drk per cubic foot low-pressure gas is, 



(L. P. Cap.) = (H. P. Cap. co\d)R p . 



lork per cubic foot cooled gas delivered is, 



(H. P. Cap. cold) s-1 

/rain, from Fig. 32, 



'hich is to say that, when best-receiver pressures are maintained, 
(L. P. Cap.) = (H. P. Cap. 



Example 1. Method of calculating Diagram, Fig. 32. 
ssumed data. 



P a =P b = 2llQ Ibs. per sq.ft. 

PC =Pa =best first-roceiver pressure =P a P g * =4330. 
P e =P/=best second-receiver pressure P^P g ^ =8830. 
P a =P h = 18,000 Ibs. per sq.ft. 

7&=5cu.ft. s = 1.4. 



(153) 



r.)# p (#/37 1 -l). . . . (154) 



=(H.P.Cap.hot)j%,~lT, .... (155) 

bnce 



fork per cubic foot hot gas delivered 

W 9 2s +1 f-1 

(H.P. Cap. ' 



130 ENGINEERING THERMODYNAMICS 

To obtain point C: 



/. V c =3 cu.ft. P c =4330 Ibs. sq.ft. 

Intermediate points B to C may be found by assuming various'pressures and finding 
the corresponding volumes as for V c . 
To obtain point D: 



F d = F 6 X^ =5 X^ =2.44 cu.ft. 
Pa 4330 

/. F d =2.44 cu.ft., P d =4330 Ibs. sq.ft. 

To obtain point E: 

T7 T . (P e \iA P e *Pa 

V e = V d + ( } , but = , 



by assumption of best-receiver pressure. 

Hence V e = 2.44-^-1.67 = 1.46 cu.ft., an P e = 8830 Ibs. sq.ft. 

Intermediate points D to E may be found by assuming various pressures and findin ; 
corresponding volumes as for V e , and succeeding points are found by similar method*; 
to these already used. 

7, = .72, P a = 18,000, 

Example 2. What will be the horse-power required to compress 100 cu.ft. of fre 
air per minute from 15 Ibs. per square inch absolute to 90 Ibs. per square inch gage in 
no-clearance, three-stage compressor if compression be adiabatic? What will be th 
work per cubic foot of (del.pr.) air hot or cold? 

From Eq. (153) work per cubic foot of (sup.pr.) air is, 



^X15X(7- 0952 -1) =4500 ft.-lbs., 

or 

4^00 vino 
H.P. for 100 cu.ft. per minute =33^ = 13 - 6 - 

From Eq. (154) work per cubic foot of (del.pr.) air cold is R p times that per cul 
foot of (sup.pr.) air, or in this case is 31,500 ft.-lbs. 

I From Eq. (156) work per cubic foot of (del.pr.) air hot is R p 3s times that per cub 
foot of (sup.pr.) air, or in this case 5.8 X45,000 =46,200 ft.-lbs. 



WORK OF COMPRESSORS 131 

Prob. 1. What work will be required to supply 2000 cu.ft. of air at 200 Ibs. per inch 
j,ge pressure if compressing is done adiabatically by three-stage compressors, taking 
; T at atmosphere, neglecting the clearances? 

Prob. 2. A motor is available for running a compressor for compressing gas, for 
lich s equals 1.3. If 60 per cent of the input of the motor can be expended on the 
,r, to what delivery pressure can a cubic foot of air at atmospheric pressure be com- 
;essed in a zero clearance three-stage machine? How many cubic feet per minute 
. uld be compressed to a pressure of 100 Ibs. gage per H.P. input to motor? 

Prob. 3. Two compressors are of the same size and speed. One is compressing 
r so that exponent is 1.4, the other a gas so that exponent is 1.1. Each is three stage, 
'hich will require the greater power to drive, and the greater power per cubic foot 
, (sup.pr.) gas, and per cubic foot of (del.pr.) gas, hot, and how much more, neglect- 
g clearance? 

Prob. 4. How will the work per cubic foot of (sup.pr.) air and per cubic foot of (del. 
K) air differ for a three-stage compressor compressing from atmosphere to 150 Ibs. per 
,uare inch gage from a single- and a two-stage, neglecting clearance? 

Prob. 5. A table in " Power " gives the steam used per hour in compressing air to 
irious pressures single stage. A value for air compressed to 100 Ibs. is 9.9 Ibs. steam 
;;r hour per 100 cu.ft. of free air. Using the same ratio of work to steam, find the value 
;r the steam if compression had been three-stage, zero clearances to be assumed. 

Prob. 6. A 5 in. drill requires 200 cu.ft. of free air per minute at 100 Ibs. per square 
ich gage pressure. What work will be required to compress air for 20 such drills if 
tree-stage compressors are used, compared to single-stage for no clearance? 

Prob. 7. What would be the steam horse-power of a compressor delivering 150 
i.ft. of air per minute at 500 Ibs. per square inch pressure if compression is three-stage, 
jliabatic, clearance zero, and mechanical efficiency of compressor 80 per cent? 

14. Three-stage Compressor with Clearance, Perfect Interceding Expo- 
untial Compression (Cycle 8), Best-receiver Pressures, Equality of Stages. 
v ork and Capacity in Terms of Pressures and Volumes. The pressure- 
T )lume diagrams of the three-stage compression is shown in Figs. 33, 34 and 
S, on which the clearance volume and displacements, low-pressure capacity 
;id high-pressure or delivery capacity for hot gas are indicated. 

If perfect intercooling exists, as is here assumed, 



id also ^ (157) 



= (V d - V l )P d = (V r - Vi)P f 
(L. P. Cap.)P 6 =(H.P. Cap. cold)P,. 
Apply Eq. (57) to the three stages and the entire work done is, 

W = ~P>(V<,-V a )[(~?^-l] (first stage) 



s r /p e \ s ~ i ] 

2jr Pd(V d Vi) ( ~ } s 1 (second stage) 

LY* <*/ 



f-,P,(V f - V,} V pT - 1 (third stage) 

o 1 



. . . (158) 



132 



ENGINEERING THERMODYNAMICS 



By use of the above conditions of perfect interceding Eq. (157) thi.< 
expression becomes, 



(159 



t 
. c 

bC c 




5 



^ooji aa^nbg aad spunoj ui saanssoaa; 



in which 



p 

c 



WORK OF COMPRESSOES 



133 



L terms of supply pressure, pounds per square inch, low-pressure capacity, 
coic foot and ratios of pressures as above, the work of a three-stage corn- 
p ;ssor with perfect intercooling and with clearance is 



= 14 (sup.pr.)(L.P. Cap.) 



(160) 




hich is identical with Eq. (141), showing that clearance has no effect upon the 
ork for a given capacity. 



134 



ENGINEERING THERMODYNAMICS 



It readily follows that the work per unit of gas is independent of clearai 
and hence Eqs. (142), (143) and (144), will give a correct value for the w[ 




f 8 
^ooj arctibg aad 'sqi 



CO 



5V 



=3 



& 



per cubic foot of gas supplied, per cubic feet delivered and cooled, and 1 
cubic foot as delivered hot, respectively. 

Since in two-stage compressors the reasoning leading to the determinat c 
of best-receiver pressure applies equally well with and without clearance, it 
since the value of best-receiver pressures for three-stage are found by cf- 



WORK OF COMPRESSORS 135 

slering the three-stage a combination of one- and two stage-compressors, the 
sue expressions for best-receiver pressures will hold with clearance as without; 
S3 Eqs. (147) and (148). 

P e =(best 2 rec.pr.) = [(sup.pr.)(del.pr.) 2 ]*. 
P c = (best 1 rec.pr.) = [(sup.pr.) 2 (del.pr.)]*. 

The use of these expressions for best-receiver pressures leads to the same 
isult as for no clearance Eq. (150), except for the volumes, 

r ork, three-stage] best-receiver's pressure with 'clearance 

8-1 

(161) 



3 -l] 



'hich is stated below in terms of supply pressure, pounds per square inch low- 
j-essure capacity, cubic foot, and ratio of compression R p , 



r ork, three-stage best-receiver pressure. 

s-l 



(162) 



'hich is identical with Eq. (151). 

From this may be obtained expressions for the work per cubic foot of low- 
ressure gas supplied to compressor per cubic foot of gas delivered and cooled, 
;id per cubic foot of gas as delivered hot from the compressor, when the re- 
viver-pressures are best, and these will be respectively identical with Eqs. 
.53), (154), and (156), in the foregoing section. 

i 15. Three-stage Compressor, any Receiver-pressure Exponential Com- 
ession. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure, 
id Horse-power in Terms of Dimensions of Cylinders and Clearances. 

- 

DI = displacement of the first-stage cylinder, r in cubic feet=(F& F OT ); 

D 2 = displacement of the second-stage cylinder, in cubic feet = (F d Ft); 
Da = displacement of the third-stage cylinder, in cubic feet = (F/ Fa). 

ci, C2, cs are the clearances of the first, second and third stages respectively, 
;ated as fractions of the displacement, so that, 

Clearance volume, 1st stage, in cubic f eet = V m 
Clearance volume, 2d stage, in cubic feet = Ft = 
Clearance volume, 3d stage, in cubic f eet =Vn = 



136 ENGINEERING THERMODYNAMICS 

The low-pressure capacity of the first stage, and hence for the compress^ 
is (F 6 Fa), and in terms of clearance, ci, and displacement DI of the fir 
stage is, according to Eq. (64), 



(L.P. 

For the second stage, the low-pressure capacity is (Va Vj) and is equal t 

i 

(L. P. C&p.2)=D 2 (l+C2-C2R P 2 *)=D 2 E V2 , .... (16- 

and for the third stage (F/ Vj) or, 

j_ 
(L. P. Cap.3)=3(l+C3-C3# P 3 7 )=>3# P 3. .... (16; 

The volumetric efficiency of 1st stage is 



Volumetric efficiency of second stage is 



Volumetric efficiency of third stage is 



The work of the three-stage compressor with the assistance of Eq. (163) ma 
be stated in terms of supply pressure, pounds per square inch, displacemen 
of first-stage cylinder, in cubic feet and volumetric efficiency of first stage, am 
also ratios of compression existing in the first, second, and third stages, 

. (16?. 

To make use of this formula for the work of the compressor the tw 
receiver pressures must be known, and it is, therefore, important to derive j 
relation between receiver pressures, displacements and clearances or volumetr: 
efficiencies. 

The assumption of perfect intercooling which has already been made U 
of in obtaining Eq. (169), regardless of the receiver-pressure, requires that- 
seeEq. (157): 

(L.P. Cap.i) (sup.pr.) = (L.P. Cap. 2 ) (1 rec.pr.) 

= (L.P. Cap. 3 )(2rec.pr.). -(170 



WOEK OF COMPRESSORS 



137 



Using values of capacities in Eqs. (163), (164), and (165) and solving for 
urst- and second-receiver pressures. 



nd 



. 
(2 rec.pr.) 



(L. P. Cap.i) 



(L. P. Cap.i) 
---. 



R p i = 



(1 rec.pr.) _D\E v i 
(sup.pr.) ~D 2 E v2 ' 

(2 rec.pr.) D 2 E v2 



p (1 rec.pr.) 
>y definition, (del.pr.) =R P (sup.pr.), 



r, (171) 



D ^. (172) 



(173) 



(174) 



(del.pr. _ (sup.pr.) _ 

"2>3 /o \ -*-* / P /r \ "i 

(2 rec.pr.) (2 rec.pr.) 



. . . (175) 



'he work of the three-stage compressor may then be stated in terms of 
supply pressure, pounds per square inch, displacements, cubic feet, volumetric 
fficiencies, and overall ratio of compression, R P} as follows : 



s-l 



In Terms of Pressures, Displacements, and Clearances, an expression can be 
'ivritten by substitution of values of E v \, E v2 and E v % from Eqs. (166), (167) and 
J168), but it becomes a long expression, further complicated by the fact that 
$ P i, R p2 and ^3 remain in it. This may be solved by the approximation 
ipased first upon the assumption that all volumetric efficiencies are equal to each 
!>ther or to unity when 



#1 

D 2 

&2 (If volumetric efficiencies are each equal to each other 
D 3 or to unity) (177) 



138 ENGINEERING THERMODYNAMICS 

This process amounts to the same thing as evaluating E v i, E V 2, and E v z fi 
Eqs. (166), (167) and (168), making use of the approximation Eq. (177) anc 
substituting the values found in Eq. (176). 

Since the above can be done with any expression which is in terms of volu 
metric efficiencies, the following formulae will be derived from Eq. (176), as i 
stands. 

The mean effective pressure of the three-stage compressor reduced to the first 
stage cylinder is found by dividing the work of the entire cycle, Eq. (176) Ir 
displacement of the first stage, and by 144 to reduce to pounds per square inch 



(m.e.p.) reduced to first stage cylinder, 
W s 






Note here that this may also be obtained by multiplying (work per cubic foe 
supplied) by (volumetric efficiency of first stage) and dividing the product by 14^ 
The indicated horse-power of a compressor performing n cycles per minul 
will be equal to the work per cycle multiplied by n and divided by 33,000, o 
for the three-stage compressor with general receiver pressures, 

o VP ^. F '- 1 /n-tf\.-i 

LH.P. = -2^ 



For n may be substituted the number of revolutions per minute, N, divid< 
by the revolutions required to complete one cycle 



N 
. 
z 



The horse-power per cubic foot of gas suppliedjper minute is 

LH.P. _s _ (sup.pr.) [ (DiE,i\ ^ , (D^} ^ 

n(L.P.Cap.) s-1 229.2 l\ ' ' 



Horse-power per cubic foot gas delivered and cooled per minute is 

I.H.P. s (del.pr.) \/DiE v i\^ L /D 2 E V , 

n(H.P.Cap.cold) s-1 229.2 \\D 2 E v2 ) r \DMt t 

81 



WORK OF COMPRESSORS 
lorse power per cubic foot hot gas delivered per minute is 



139 



I.H.P. 



_ s (sup.pr.) 
nCH.P.Cap.hot)**-! 229.2 



8-1 



"* 



L ["/>!#, A 
* |_ \BJt*J 



:Fhe last equation is obtained by means of the relation 



(L. P. Cap.) = (H. P. Cap. hot) X 



Vint^ V 

. not; X 



X 



.pr. \ sup.pr 






(183) 



// clearance is zero or negligible, these expressions may be rewritten, putting 
&>, ^2 and E'vs each equal to unity. 






B 



H.P. per cubic foot of gas supplied per minute is 



n(L.P.Cap.)- 229.2 
H.P. per cubic foot delivered and cooled per minute is 






, . 



I.H.P. _ (del, pr.) [7>i 



s-l 



s-l 



pr.)r/J) 1 
.2 [\5i 









n(H.P.Cap.cold)~ 229 
3. P. per cubic foot hot gas delivered per minute is 

I.H.P. g (sup.pr.) /DA ^ if/ftX" , / 

n(H.P. Cap.hot) s-l 229.2 \Z) 3 / "l\ft 



s-l 

8T-] 









140 



ENGINEERING THERMODYNAMICS 



Example 1. Method of calculating Diagram, Fig. 35. 
Assumed data: 

P a = Pb =2116 Ibs. per square foot. 
P c =p d =P l =P m =4330 Ibs. per square foot. 
P e =P f = Pj=P t =S830 Ibs. per square foot. 

P g =p h = 18,000 Ibs. per square foot. 
d=7.5 per cent for all cylinders; s = 1.4. 
L.P. capacity 5 cu.ft. 

To obtain point M: 



From formula Eq. (163) L. P. Capi. = A(l +Ci -CiR p i ) 

5 =Di(l +.075 -.075x1.67) or Z>i= 5.3 cu.ft. and clearance volume 
F m =5.3x.075=.387cu.ft. 



or 



V m = .39 cu.ft. ; P m =4330 Ibs. sq.ft. ; 



~ i- 4 . =.39X1.67 = .67 cu.ft. 



Therefore, 



To obtain Point A : 



Additional points M to A may be found by assuming pressures and finding cori 
spending volumes as for F a . 

To obtain point B : 
Therefore, 
To obtain point C: 



. P. Capi.) =.67+5 =5.67 cu.ft. 
F 6 =5.67 cu.ft.; P b =2116 Ibs. sq. ft.; 






- =5.67-5-1.67 =3.45 cu.ft. 



Therefore, 



V c = 3.45 cu.ft. ; P c = 4330 Ibs. sq.ft. 



Intermediate points B to C may be found by assuming various pressures and find] 
corresponding volumes as for F&. 



WORK OF COMPRESSORS 141* 



obtain point D : 



Volume at D is the displacement plus clearance of the intermediate cylinder. This 
cmot be found until the capacity is known. Applying the same sort of relations as 
\? re used in calculating the diagram for the two-stage case with clearance, 

1 

D 2 (l+C2-C 2 Rp2) =2.44 or D 2 =2.57, 

id clearance volume. 

V k = . 075X2.57-. 192 cu.ft., 
tnce, 

V d =2.57 +.19 =2.76 cu.ft. 
lierefore, 

V d =2.76 cu.ft; P d =4330 Ibs. sq.ft.; 

The rest of the points are determined by methods that require no further explana- 
t<n and as pressures were fixed only volumes are to be found. These have the following 
^lues, which should be checked: 

1=1.65; 7, = 1.32; 7, = .79; 7*=.09; 7, = .15; 7, = .32; 7* = .65; 7,, = 1.23; V z = . 14. 

Example 2. A three-stage compressor is compressing air from atmosphere to 140 
k per square inch absolute. The low-pressure cylinder is 32x24 ins. and is known to 
Ive a clearance of 5 per cent. From gages on the machine it is noted that the first- 
Reiver pressure is 15 Ibs. per square inch gage and the second-receiver pressure is 
i Ibs. per square inch gage. What horse-power is being developed if the speed is 

R.P'.M. and s = 1.4? From the formula Eq. (169), 

9 r 8-1 Sj-J. s-l 

W = 144--(sup.pr.)D 1 # Pl [# pl +R P2 s +R PS , 

1 From gage readings 

7? 3 2 7? 7 2^- 7? 14 2 

^ = ^=2. ^ 2=3o= 2.33, R P3 - -2. 

i 
^ n (l+Ci-Cifti7) fromEq. (166), 

E n = (l +.05 -.05X1. 65) =67.5 per cent, 
f 'ence, 

W = 144x-^ -Xl5xll.2x.675(1.22+1.28 + 1.22-3); 

= 59,200 ft.lbs. per stroke or 200 X59,200 ft. = Ibs. per minute; 
= 358 I.H.P. 



142 ENGINEERING THERMODYNAMICS 

Examples. Another compressor has cylinders 12x20x32x24 in. and it is know 
that the volumetric efficiencies of the high, intermediate and low-pressure cylinders ai 
respectively 70 per cent, 85 per cent and 98 per cent. The (del.pr.) is 150 Ibs. per squar 
inch absolute. What is the horse-power in this case if the speed is 100 R.P.M.? 

From the formula Eq. (176), 



= (1.309+1.495+1 -3) =66,400 ft.-lbs. per stroke, 

200X66,400 
Whence I.H.P. = -33^5- =402. 

Prob. 1. What will be the horse-power required to drive a 12 X22 X34 X30 in. th 
stage compressor with volumetric efficiencies of 75, 85, and 95 per cent in the high 
intermediate and low-pressure cylinders, at 100 R.P.M. when compressing natural ga 
from 25 Ibs. per square inch gage to 300 Ibs. per square inch gage, adiabatically? 

Prob. 2. A three-stage compressor for supplying air for a compressed-air locomc 
tive receives air at atmosphere and delivers it at 800 Ibs. per square inch gage. Shoul 
the receiver pressures be 50 Ibs. and 220 Ibs. respectively in the first and second and th 
volumetric efficiency of the first stage 90 per cent, what would be its displacemei} 
and horse-power when compressing 125 cu.ft. of free air per minute, adiabatically 
What are the cylinder displacements? 

Prob. 3. Find the work done on a gas, the value for s of which is 1.3, in compressiD 
it from atmosphere to 7 atmospheres absolute, adiabatically in three stages, the com 
pressor having a low-pressure cylinder displacement of 60 cu.ft. per minute and a voli 
metric efficiency of 95 per cent, first receiver pressure being 2 atmospheres absolut* 
and second-receiver pressure 4 atmospheres absolute. If air were being compresse 
instead of the above gas, how would the work vary? 

Prob. 4. The cylinders of a compressor are 8x12x18x24 ins. and clearance suo 
as to give volumetric efficiencies of 80, 90 and 98 per cent in the different cylinders in tl: 
order given. Compressor is double acting, running at 120 R.P.M. and compressing ai 
adiabatically from 14 Ibs. per square inch absolute to 150 Ibs. per square inch gag( 
What is the capacity in cubic feet per minute, work done per cubic feet of (sup.pr.) aii 
(del.pr.) air hot and cold and the horse-power of the compressor? .What would t 
the effect on these quantities if the clearances were neglected? 

Prob. 5. If the cylinders of a compressor are 10x14x20x18 ins., and clearanc* 
are 8, 5 and 3 per cent, what work is being done in adiabatically compressing air fro: 
10 Ibs. per square inch absolute to 100 bs. per square inch gage? 

NOTE: Solve by approximate method. 

Prob. 6. For special reasons it is planned to keep the first-receiver pressure of 
three-stage compressor at 30 Ibs. per square inch absolute, the second-receiver pressure : 
60 Ibs. per square inch absolute, and the line pressure at 120 Ibs. per square inch absolu 




WOEK OF COMPRESSORS 143 

le (sup.pr.) is 14 Ibs. per square inch absolute. If the clearances are 4 per cent in 
t: low and 8 per cent in the intermediate and high-pressure cylinders, what must be 
t ! cylinder sizes to handle 500 cu.ft. of free air per minute at 120 R.P.M. and what 
p ^er must be supplied to the compressor on a basis of 80 per cent mechanical effi- 
cacy, for a value of s equal to 1.39? 

Prob. 7. Should the above pressures (Prob. 6) be gage pressures instead of absolute, 
r N would the quantities to be found be affected? 

Prob. 8. The receiver pressures on a C0 2 gas compressor are 50 Ibs. per square inch 
r solute, and 200 Ibs per square inch absolute, the (del.pr.) being 1000 Ibs. per square 
i h absolute. The mach ne has a low-pressure cylinder 8x10 ins. with 3 per cent 
carance. What horse-power will be required to run it at 100 R.P.M. and what would 
1 the resultant horse-power and capacity if each pressure were halved? (Sup.pr.) = 14.7 
1 . per square inch. 

16. Three-stage Compressor with Best-receiver Pressures Exponential 
(mpression. Capacity, Volumetric Efficiency, Work, Mean Effective Pressure 
ad Horse-power in Terms of Dimensions of Cylinders and Clearances. It 

vs found that for the three-stage adiabatic compressor with perfect inter- 
CDling, the work was a minimum if the first and second receivers had pressures 
cfined as follows, see Eqs. (147) and (148) : 

(best 1 rec.pr.) = [(sup.pr.) 2 (del.pr.)]*. ..... (188) 

(best 2 rec.pr.) = [(sup.pr.) (del.pr.) 2 ]*. ..... (189) 

, _ (best 1 rec.pr.) _ / del.pr.X* _ 
pl ~ (sup.pr.) 



(best 2 rec.pr.) /del.prA* 

2 = 7i i = K>J> .... 

(best 1 rec.pr.) \sup.pr./ 



(del.pr.) /deLprA* 

= 7r~^^ \ = \ I = 

(best 2 rec.pr.) V sup.pr./ 



The use of these values in connection with expressions previously given 
f< volumetric efficiency, Eqs. (166), (167) and (168), gives, 

\lumetricefficiencyoffirststage =E c i = (l+c 1 -c 1 R/~*) .... (193) 
Ylumetric efficiency of second stage = #2 = (1+ C 2 C 2#? 3s ) .... (149) 
V.lumetric efficiency of third stage =^ P 3 = (l+c 3 c 3 R P 3s ) .... (195) 



144 



ENGINEERING THERMODYNAMICS 



The work of the three-stage compressor with best-receiver pressures, E( 
(162), when expressed in terms of displacement and volumetric efficiency beconu 



s 1 



s 



where 



(del.pr.) 

7 ^ 

(sup.pr.) 



If clearance is known, the value of E,\ may be ascertained by Eq. (19; 
and inserted in Eq. (196). Since this may be so readily done the substitutic 
will not here be made. 

The mean effective pressure of the compressor referred to the first stage 
obtained by dividing the work Eq. (196) by 144 DI: 

(m.e.p.) referred to first-stage cylinder 



W 



-1 



(19 



The mean effective pressures of the respective stages, due to the equali 
of work done in the three stages will be as follows: 



For first stage 



(m.e.p.) = -(sup.pr.)^! (fl/ST-l) 

o 1 



(K 



For second stage 



(m.e.p.) = 



- 1) 



For third stage 



(m.e.p.) = ^ (sup.pr.)^.i (72* V- 1). 



But 



.prO'* = (1 rec.pr.) = [(sup.pr.) 2 (del.pr.)]*, 



and also 



.pr.) = (2 rec.pr.) = [(sup.pr.) (del.pr.) 2 ]*. 




WORK OF COMPRESSORS 



lence 

'or second stage 



s-1 



(m.e.p.) =-:[(sup.pr.) 2 (deLpr.)]^2(^ -1). 



i. 



third stage 






145 



(201) 



(202) 



Conditions to Give Best-receiver Pressures. All the foregoing discussion 
;>f best-receiver pressures for the three-stage compressor can apply only to cases 
;Q which all the conditions are fulfilled necessary to the existence of best-receiver 
oressures. These conditions are expressed by equations (173), (174), (175), 
190), (191), and (192), which may be combined as follows: 



(1) (2) (3) (4) 

(L. P. Cap.i; = (L. P. Cap. 2 ) = DiE vl = 



(L.P.Cap.2) (L.P.Cap.3) 



(5) 



(6) 



. . (203) 



'arts (1) and (2) of this equation state the requirements in terms of 
capacities; (3) and (4) in terms of displacements and volumetric efficiencies; 
.'5) and (6) in terms of displacements and clearances. In order, then, that best- 
Deceiver pressure may be obtained, there must be a certain relation between 
:he given ratio of compression and dimensions of cylinders and clearances. 
Since, after the compressor is once built these dimensions are fixed, a given 
nulti-stage compressor can be made to give best-receiver pressures only when 
compressing through a given range, i.e., when R p has a definite value. If R p 
has any other value the receiver pressures are not best, and the methods of the 
previous Section (15) must be applied. 

When clearance percentages are equal in all three cylinders, ci=C2=cs, and 
the volumetric efficiencies are all equal then, when best-receiver pressures exist, 
Eq. (203) becomes, 

RJ j^ = j^ = for equal clearance per cent. . . (204) 



Evidently this same expression holds if clearances are all zero or negligible. 
What constitutes negligible clearance is a question requiring careful thought 
and is dependent upon the ratio of compression and the percentage of error 
allowable. 



146 ENGINEERING THERMODYNAMICS 

Indicated horse-power of the compressor is found by multiplying the wor 
per cycle, Eq. (196) by the number of cycles per minute, n, and dividing th 
product by 33,000. 



-l) .... (20fi 



From this are obtained the following: 
H.P. per cubic foot supplied per minute 

I.H.P. s (sup.pr.) , ii^ 1 



H.P. per cubic foot delivered and cooled per minute 

I.H.P. s (del.pr.) fzi 1 

n(H. P. Cap. cold) = 7-^l ~J&T (Rp * 1} ..... (20 ' 

H.P. per cubic foot delivered hot per minute 



(sup.pr.) 2 -il =! 

=-: K v 3s (H p 3s 1). . . (ZOl 



(See Eq. (156)). 



It is useful to note that these expressions are all independent of clearance 
which is to be expected, since the multi-stage compressor may be regarded as 
series of single-stage compressors, and in single stage such an independenc 
was found for work and horse-power per unit of capacity. 

Example. If the following three-stage compressor be run at best-receiver pressure 
what will be the horse-power and the best-receiver pressures? Compressor has lov 
pressure cylinder 32x24 ins. with 5 per cent clearance, is compressing air from atmo>| 
phere to 140 Ibs. pier square inch absolute, so that s equals 1.4 and it runs at 100 R.P.& 

From the formula Eq.~(196) 

From the formula Eq. (188) 

(best 1 rec.pr.)= [(sup.pr.) 2 (del.pr.)]i 
= (15) 2 X140]*=31.6. 



WORK OF COMPRESSORS 147 

From Eq. (189) 

(best 2 rec.pr.) =[ (sup.pr.) (del.pr.) 2 ]* 

= [15X(140) 2 ]4=66.5. 
From Eq. (193) 

i 



;mco, 

W =432 X^ X15 xll.2 X96.5 X (9.35 95 -l) =59,000 ft.-lbs., 



, p 59,000X200 
33,000 

Prob. 1. There is available for running a compressor 175 H.P. How many cu.ft. 
' free air per minute can be compressed from atmosphere to 150 Ibs. per square inch 
ige by a three-stage adiabatic compressor with best-receiver pressures? 

Prob. 2. The low-pressure cylinder of a three-stage compressor has a capacity of 
i cu.ft. per stroke. If the stroke of all three cylinders is 18 ins., what must be the 
ameters of the intermediate and high to insure best-receiver pressures, if clearance 
.5 neglected, and (sup.pr.) be 1 Ib.jper square inch absolute and (del.pr.) 15 Ibs. per 
luare inch absolute, s being 1.4. 

Prob. 3. The above compressor is used as a dry-vacuum pump for use with a sur- 
ce condenser. If 800 cu.ft. of (sup.pr.) gas must be handled per minute what horse- 
>wer will be needed to run it? What will be the horse-power per cubic foot of atmos- 
jieric air? 

Prob. 4. Will a 15 X22 X34 X24 in. compressor with clearances of 3, 5 and 8 per cent 

low, intermediate and high-pressure cylinders respectively be working at best-receiver 

ressures when (sup.pr.) is 15 Ibs. per square inch absolute and (del.pr.) 150 Ibs. per 

jiuare inch absolute? If not, find by trial, the approximate (del.pr.), for which this 

achine is best, with s equal to 1.4? 

Prob. 5. For the best (del.pr.) as found above find the horse-power to run the 
achine at 100 R.P.M. and also the horse-power per cu.ft. of (del. pr.) air cold? 

Prob. 6. Should this compressor be used for compressing ammon'a would tl.c 
')st (del. pr.) change, and if so what would be its value? Also what power would Le 
;eded for this case? 

Prob. 7. Compare the work necessary to compress adiabatically in three stages from 
') Ibs. per square inch absolute to 200 Ibs. per square inch absolute, the following gases: 

Air; Oxygen; Gas-engine mixtures, for which s = 1.36. 

Prob. 8. For 5 per cent clearance in all the cylinders what must be the cylinder ratio 
:r best-receiver pressure and a pressure ratio of 10? 



148 ENGINEEEING THERMODYNAMICS 

Prob. 9. A compressor, the low-pressure cylinder of which is 30x20 ins. with 5 pi 
cent clearance is compressing air adiabatically from atmosphere to 150 Ibs. per squa: 
inch gage, at best-receiver pressure. Due to a sudden demand for air the (del. pr 
drops to 100 Ibs. per square inch gage. Assuming that the (1 rec. pr.) dropped to 5 lb 
per square inch gage and (2 rec.pr.) dropped to 40 Ibs. per square inch gage, how muc 
would the speed rise if the power supplied to machine was not changed? 

17. Comparative Economy or Efficiency of Compressors. As the prin 
duty of compressors of all sorts is to move gas or vapor from a region of IQI 
to a region of high pressure, and as this process always requires the expenditui 
of work, the compressor process which is most economical is the one thi 
accomplishes the desired transference with the least work. In this sense, thei 
economy of compression means something different than efficiency, as ord 
narily considered. Ordinarily, efficiency is the ratio of the energy at one poii 
in a train of transmission or transformation, to the energy at another poin 
whereas with compressors, economy of compression is understood to mean tl: 
ratio of the work required to compress and deliver a unit of gas, moving 
from a low- ,to a high-pressure place, to the work that would have been require 
by some other process or hypothesis, referred to as a standard. This econom 
of compression must not be confused with efficiency of compressors as machine 
as it is merely a comparison of the work in the compressor cylinder for an actu; 
case or hypothesis to that for some other hypothesis taken as a standard. Tl 
standard of comparison may be any one of several possible, and unfortunate! 
there is no accepted practice with regard to this standard. It will, therefor 
be necessary to specify the standard of reference whenever economy of compre 
sion is under consideration. The following standards have been used wit 
some propriety and each is as useful, as it supplies the sort of information reall 
desired. 

First Standard. The work per cubic foot of supply gas necessary to COD 
press isothermally (Cycle 1), from the supply pressure to the delivery pressu 
of the existing compressor and to deliver at the high pressure is less than that 
any commercial process of compression, and may be taken as a standard f 
comparison. Since, however, actual compressors never depart greatly fro* 
the adiabatic law, their economy compared with the isothermal standard w/| 
always be low, making their performance seem poor, whereas they may be ' 
nearly perfect as is possible, so that it may appear that some other standa 
would be a better indication of their excellence. 

Second Standard. The work per cubic foot of gas supplied when compresst 
adiabatically in a single stage (Cycle 3), if taken as a standard, will indicate 
high economy, near unity for single-stage compressors, and an economy abo 
unity for most multi-stage compressors. For the purpose of comparison 
will be equally as good as the first standard, and the excess of the economy ov 
unity will be a measure of the saving over single-stage adiabatic compressic 
Since, however, single-stage adiabatic compression is not the most economic 
obtainable in practice for many cases, this standard may give an incorrc 
idea of the perfection of the compressor. 



WORK OF COMPRESSORS 149 

Third Standard. Due to the facts noted above, it may be a better indica- 
on of the degree of perfection of the compressor to compare the work per 
ibic foot of gas supplied with that computed for the standard adiabatic cycle 
ost nearly approaching that of the compressor. This standard is, however, 
)en to the objection that a multi-stage compressor is not referred to the same 
ycle as a single-stage compressor, and a multi-stage compressor with other 
lan best-receiver pressure is not referred to the same cycle as another operating 
; ith best-receiver pressure. This is, therefore, not a desirable standard for 
omparing compressors of different types with one another, although it doe s 
low to what extent the compressor approaches the hypothetical best condi- 
,on for its own type and size. 

Other standards might be chosen for special reasons, each having a value 
i proportion as it supplies the information that is sought. 

It is seen from the discussion of the second standard that its only advantage 
;ver the first is in that it affords a measure of the saving or loss as compared 
4th the single-stage adiabatic compressor cycle. 

If the first standard, that of the isothermal compressor cycle, be adopted 
3r the purpose of comparison, it at once gives a measure of comparison with 
he isothermal, which is more "and more nearly approached as the number 
f stages is increased, though never quite reached, or as the gas is more effect- 
yely cooled during compression. It may be regarded as the limiting case of 
lulti-stage compression with perfect intercooling, or the limiting case of con- 
inuous cooling. , 

In order to ascertain how nearly the actual compressor approaches the 
Adiabatic cycle most nearly representing its working conditions, the economy of 
f the various reference cycles heretofore discussed may be tabulated or charted, 
,nd the economy of the cycle as compared with that of the actual performance 
>f the compressor will give the required information. The process of com- 
mtation by which this information is obtained will depend upon the nature 
j)f information sought. The economy of actual compressor compared with the 
isothermal may be stated in any of the following ways: 



Computed work per cubic foot supplied, isothermal ( . 
Indicated work per cu.ft. actual gas supplied to compressor 

I.H.P. per cubic foot per minute supplied, isothermal 
I.H.P. per cubic foot per minute actual supplied 

Single stage 

(209) 
(m.e.p.) isothermal, pounds per square inch, no clearance 

. _ _ ( {*} 

(m.e.p.) actual -r- true volumetric efficiency 
Multi-stage 

(m.e.p.) isothermal, no clearance 



(m.e.p.) reduced to first stage 4- first stage vol. eff. 



150 ENGINEERING THERMODYNAMICS 

In this connection it is useful to note that for the case of the no-clearan 
cycles, the work per cubic foot of supply is equal to the mean effective pressu 
(M.E.P.) in pounds per square foot, and when divided by 144 gives (m.e.j 
in pounds per square inch. Also, that in cases with clearance, or even actu 
compressors with negligible clearance, but in which, due to leakage and oth 
causes, the true volumetric efficiency is not equal to unity, 

Work per cubic foot gas supplied X E v =1 44 (m.e. p.). . . (21 

The information that is ordinarily available to determine the econon 
of the compressor will be in the form of indicator cards from which the (m.e.f 
for the individual cylinders may be obtained with ordinary accuracy. TJ 
volumetric efficiency may be approximated from the indicator cards also, b 
with certain errors due to leakage and heating, that will be discussi 
later. If by this or other more accurate means the true volumetr 
efficiency is found, the information required for the use of Eq. (209) ( 
or (d) is available. Evaluation of the numerator may be had by Eq. (31 
which is repeated below, or by reference to the curve sheets found at the ei 
of this chapter. (Fig. 50.) 

Mean effective pressure, in pounds per square inch for the isothermal coi 
pressor without clearance is given by 

(m.e.p.) isothermal =(sup.pr.) log e R P (21 

The curve sheet mentioned above also gives the economy of adiabat 
cycles of single stage, also two and three stages with best-receiver pressure 
The value of s will depend upon the substance compressed and its conditio 
The curve sheet is arranged to give the choice of the proper value of s applyii 
to the specific problem. 

If it is required to find the economy of an actual compressor referred 
the third standard, i.e., that hypothetical adiabatic cycle which most near 
approaches the actual, then 



Economy by third standard is 

Econ. actual referred to isothermal 



Econ. hypothetical referred to isothermal* 



. . (21 



It is important to notice that for a vapor an isothermal process is not o<j 
following the law PXV = constant. What has, in this section, been called 
isothermal is correctly so called only so long as the substance is a gas. Sini: 
however, the pressure-volume analysis is not adequate for the treatment 
vapors, and as they will be discussed under the subject of Heat and Woi 
Chapter VI, it is best to regard this section as referring only to the treatmc 
of gases, or superheated vapors which act very nearly as gases. Howev 
it must be understood that whenever the curve follows the law PXF = consta: 
the isothermal equations for work apply, even if the substance be a vaf 
and the process is not isothermal. 






WORK OF COMPRESSORS 151 



18. Conditions of Maximum Work of Compressors. Certain types of com- 
ressors are intended to operate with a delivery pressure approximately con- 
tant, but may have a varying supply pressure. Such a case is found in pumps 
r compressors intended to create or maintain a vacuum and in pumping 
tatural gas from wells to pipe lines. The former deliver to atmosphere, thus 
laving a substantially constant delivery pressure. The supply pressure, 
lowever, is variable, depending upon the vacuum maintained. In order that 
uch a compressor may have supplied to it a sufficient amount of power to 
;eep it running under all conditions, it is desirable to learn in what way this 
>ower required will vary, and if it reaches a maximum what is its value, and 
,inder what conditions. 

Examine first the expression for work of a single-stage adiabatic compressor 
yith clearance. The work per cycle will vary directly as the mean effective 
Pressure. Eq. (69.) 

(213) 



sup.pr. sup.pr 

This will have a maximum value when 



^(sup.pr.) 
>r when 

/deLprA^ , f 1+c _^/deLprAri 

\sup.pr./ 1-f-c L s \sup.pr./ J 

Solving this for the value of supply pressure will give that supply pressure 
it which the work will be a maximum, in terms of a given delivery pressure, 
clearance and the exponent s. 

The assumption most commonly used is that clearance is zero. If this is 
rue~or the assumption permissible, the above equation becomes simplified, 



(sup.pr. 



(215) 



Che value of s for air, for instance, is 1.406, and hence the ratio of compression 
i'or maximum work for the hypothetical air compressor is 



3 ' 46 



(1.406)' = 3.26 ......... . (216) 

It may be noted that when s = 1 in the above expression, the value of the 
;atio of compression become indeterminate. To find the supply pressure for 
naximum work in this case, take the expression for mean effective pressure 
'or the isothermal compressor (s = l), Eq. (43), 

. . . ( 2 17) 



152 ENGINEEKING THERMODYNAMICS 

Differentiate with respect to (sup.pr.) and place the differential coefficient equa 
to zero. This process results in the expression 



loge - + _ = 

' Vsup.pr./ 1-fc Vsup.pr. 

When c=0, this becomes, 

/deLpr.X l or 
Vsup.pr./ 



The expressions Eqs. (215) and (219) given are easily solved, but Eqs. (214 
and (218) are not, and to facilitate computations requiring their solution 
the results of the computation are given graphically on the chart, Fig. 48, a 
the end of this chapter. 

The mean effective pressure for a compressor operating under maximun 
work conditions may be found by substituting the proper ratio of compressior 
found as above, in Eq. (213) or (217). In Fig. 48 are found also the result 
of this computation in the form of curves. Note in these curves that the mea: 
effective pressure is expressed as a decimal fraction of the delivery pressure 

The discussion so far applies to only single-stage compressors. The probler 
of maximum work for multi-stage compressors is somewhat different, and it 
solution is not so frequently required. Moreover, if the assumption of perfec 
intercooling is made, the results are not of great value, as a still greater amoun 
of power might be required, due to the failure for a period of time of the suppl; 
of cooling water. Consider this case first. 

If intercooling be discontinued in a multi-stage compressor, the volum 
entering the second stage will equal that delivered from the first, and similar! 
for the third and second stages. The entire work done in all stages will be thi 
same as if it had all been done in a single stage. It might be questioned a 
to whether this would hold, when the ratio of compression is much less tha:i 
designed. The first stage will compress until the volume has become as sma; 
as the low-pressure capacity of the second stage. If the delivery pressure i^ 
reached before this volume is reached, there is no work left to be done in th 
second or any subsequent stages, and, due to the pressure of the gas, the valves 
if automatic, will be lifted in the second and higher stages, and the gas m} 
be blown through, with only friction work. It appears then that under till 
condition of no intercooling the multi-stage compressor acts the same as i 
single stage, and the conditions of maximum work will be the same. 

If intercooling is maintained perfect there will still be a range of pressure 
on which all the work of compression is done in the first stage, merely blowing 
the discharge through receivers, valves and cylinders in the upper stages. 1 
this range is such that this continues beyond a ratio of pressures, which gives 
maximum (m.e.p.) for the single stage, then the maximum will have been reache-j 
while the compressor is operating single stage, and the single-stage fonnuk 
and curves may be applied to this case also. 




WOEK OF COMPRESSORS 



153 



That this condition frequently exists with multi-stage compressors of 
c linary design is shown by the fact that the ratio of compression in each stage 
i seldom less than 3, and more frequently 3.5, 4 or even more. The ratio 
c compression giving maximum work for single stage, has values from 2.5 
t 3.26, dependent on clearance and the value of 6' for the gas compressed, and 
i therefore, less for the majority of cases. 

, If a curve be drawn, Fig. 36, with ratio of compression as abscissas and 
(i.e.p.-^del.pr.) as ordinates, so long as the action is single stage, a smooth curve 
vll result, but when the ratio of compression is reached above which the second 
(Under begins to act, the curve changes direction suddenly, falling as the ratio 



A2 



.40 



.38 



7 



Max for Two Stage Cpmpressor 
Cylinder Ratic 



Single 



Stage 



Values of R P . 



36. Curve of Relation between Mean Effective Pressure per Pound of Delivery Pressure 
in Terms of Pressure Ratio for Air, showing Maximum Value. 



Decompression increases. Hence, if the ratio of cylinders is such that the single- 
* go maximum is not reached before the second stage begins to operate, the highest 
pint of the curve, or maximum work for a given delivery pressure will occur 
v>en the ratio of supply and delivery pressures is such as to make first-receiver 
p'ssure equal to delivery pressure. 

19. Actual Compressor Characteristics. Air or gas compressors are very 
cnmonly made double acting, so that for a single cylinder, two cycles will 
b performed during one revolution, one in each end of the cylinder. If a rod 
e;ends through one of these spaces and not through the other, the displace- 
n nt of that end of the cylinder will be less than the other by a volume equal 
tt. the area of rod multiplied by the stroke. To avoid mechanical shock 
I'the end of either stroke, it is necessary to leave some space between the 



154 ENGINEERING THERMODYNAMICS 

piston and cylinder head. Passages must also be provided, communicahj 
with inlet and discharge valves. The total volume remaining in this sp 
and in the passages when the piston is at the nearer end of its stroke constiti^ 
the clearance. The amount of this clearance volume varies from .5 or .($ 
one per cent in some very large compressors to as much as 4 or 5 per ceni 
the volume of displacement in good small cylinders. 

In order to study the performance of an actual compressor and to comj- 
it with the hypothetical cycle, it is necessary to obtain an indicator card, :i 
knowing the clearance and barometric pressure to convert the indicator C'< 
into a pressure volume diagram, by methods explained in Chapter I. Fig. 7 
is such a diagram for a single-stage compressor. In the pipe leading to .1 

-H.P.Cap Hot (Apparent) 




(Sup. Pr.) 



Vol. 
FIG. 37. Compressor Indicator Card Illustrating Departure from Reference Cycle. 



intake valve the pressure is determined and a horizontal line AB is dr/i 
on the diagram at a height to represent the supply pressure. Simik -v 
discharge pressure is determined and drawn on the diagram, KE. Cons ; 
the four phases of this diagram in succession. 

1. Intake Line. At a point somewhat below A the intake valve op is 
say at the point F. This remains open till a point H is reached at or near 
end of the stroke. The line connecting these two points indicates varial iv 
of pressures and volumes throughout the supply stroke. In general this 
will lie below the supply-pressure line AB due to first, the pressure necesft 
to lift the inlet valve from its seat against its spring and inertia, if autom 
and support it, and second, friction in the passages leading to the cylinder 1 
the point where the supply pressure was measured. While the former is m b 
constant the latter varies, depending upon the velocity of gases in the pass* * 
The piston attains its highest velocity near the middle of the stroke, u: 




WORK OF COMPRESSORS 155 

(using the intake line to drop below the supply pressure more at this part 
c stroke. These considerations do not, however, account completely for the 
irm of the intake line. Frequently the first portion of the line lies lower than 
1e last portion, even at points where the piston velocities are equal. This 
i more prominent on a compressor having a long supply pipe, and is due to 
is forces required to accelerate the air in the supply pipe while piston velocity 
i increasing, and to retard it while piston velocity is decreasing. In com- 
jessors where the inlet valve is mechanically operated and the supply pipe 
hg, it is possible to obtain a pressure at the end of the intake, line H, even 
i excess of the supply pressure. The effect of this upon volumetric efficiency 
\11 be noted later. 

The apparent fluctuations in pressure during the first part of the intake 
1 e may be attributed, first, to inertia vibrations of the indicator arm, in which 
.se the fluctations may not indicate real variations of pressure; second, the 
iiicator card may show true variations of pressure due to inertia of the gases 
i the supply pipe, since a moment before the valve opened at F the gases were 
sitionary in the supply pipe. When F is reached the piston is already in motion 
sd a very considerable velocity is demanded in the supply pipe to supply 
t.3 demand. This sudden acceleration can be caused only by a difference in 
pssure, which is seen to exist below and to the right of F on the diagram, 
ne suddenness of this acceleration may start a surging action which will cause 
re and fall of pressure to a decreasing, extent immediately after. A third cause 
i possible, that is, a vibration of the inlet valve due to its sudden opening 
\ien it is of the common form, mechanical valves change the conditions. It 
iclosed by weight or a spring and opened by the pressure difference. Between 
tese forces the valve disk may vibrate, so affecting the pressure. 

2. Compression Line. From the time the inlet valve closes at the point 
/ until the discharge valve opens at the point G, the gases within the cylinder 
ai being compressed. The compression is very nearly adiabatic in ordinary 
jactice, but due to the exchange of heat between the cylinder walls, at first 
fun walls and later from gas to walls, which are cooled by water jacket to 
Invent the metal from overheating, there is a slight departure from the adia- 
1 tic law almost too small to measure. 

> j A second factor which influences the form of this curve to a greater extent 
Ueakage. This may occur around the piston, permitting gas to escape from 
qe end of the cylinder to the other. During the compression process there is 
fut an excess of pressure in the other end of the cylinder due to reexpansion, 
tiding to increase pressures on the first part of compression. Later, the 
pssure rises and the pressure on the other side of the piston falls to supply 
p3ssure. During this period leakage past piston tends to decrease successive 
Fissures or lowers the compression line. Leakage also occurs through either 
c|;charge or inlet valves. The former will raise the compression line, while 
e3essive leakage of the inlet valve will lower it. 

It is then evident that unless the nature of the leakage is known, it is 
[(possible to predict the way in which it will change this line. It is, however, 



!56 ENGINEERING THERMODYNAMICS 



WORK OF COMPRESSORS 



157 



[valve. After reexpansion is completed the intake valve opens and gas enters 
the end of the cylinder under consideration. At the same time compression 
I is taking place in the other end, and later deliver}'. During these processes 
j whatever gas leaks past the piston tends to fill the end of the cylinder in which 
intake is going on. Leakage past the discharge valve also tends to fill the cylin- 
der with leakage gas. Both of these tendencies decrease the quantity of gas 
! entering through this intake valve, and its true amount when reduced to external 
; supply pressure and temperature is, therefore, less than the volume AB. 

The true low-pressure capacity of the compressor is the true volume of gas 
I under external supply conditions that enters the cylinder for each cycle. This 
i cannot be determined from the indicator card except by making certain assump- 
tions which involve some error at best. It can, however, be ascertained by 
means of additional apparatus, such as meters or calibrated nozzles or receivers, 
by means of which the true amount of gas compressed per unit of time is made 
known. This reduced to the volume per cycle under supply pressure and tem- 
perature will give the true low-pressure capacity. 

Volumetric efficiency is defined as being the ratio of low-pressure capacity 
! to displacement. On the diagram, Fig. 37, the displacement is represented 
Ito the volume scale by the horizontal distance between verticals through the 
extreme ends of the diagram, K and H. Since there are three ways in which 
the low-pressure capacity may be approximated or determined, there is a 
corresponding number of expressions for volumetric efficiency. 

1. The volumetric efficiency of the hypothetical cycle is 



^(hypothetical) = 



(hypothetical L. P. Cap.) 
(displacement) 



. .- (220) 



and this is evaluated and used in computations in the foregoing sections of 
I this chapter. 

2. The apparent volumetric efficiency is 



(apparent L. P. Cap.) / 991 x 

= -- (22. 



, N 

R(apparent) 



and would be very nearly equal to the true volumetric efficiency were it not 
for leakage valve resistance and heating during suction, but due to this may 
be very different from it. 

3. The true volumetric efficiency is 



(true L. P. Cap.) 
(displacement ) ' 



(222) 



In problems of design or prediction it is necessary either to find dimensions, 
F( speeds and power necessary to give certain actual results, or with given dimer 
* ions and speeds to ascertain the probable power and capacity or 



15 s ENGINEERING THERMODYNAMICS 

characteristics of actual performance. Since it is impossible to obtain actual 
performance identical with the hypothetical, and since the former cannot be 
computed, the most satisfactory method of estimate is to perform the computa- 
tions on the hypothetical cycle, as is explained in previous sections of this chap- 
ter, and then to apply to these results factors which have been found by 
comparing actual with hypothetical performance on existing machines as nearly 
like that under discussion as can be obtained. This necessitates access to data 
on tests performed on compressors in which not only indicator cards are taken 
and speed recorded, but also some reliable measurement of gas compressed. 
The following* factors or ratios will be found of much use, and should be 
evaluated whenever such data is to be had : 

# r (true)_ (true L. P. Cap.) 

61 ~ E, (hypothetical) (hypothetical L. P. Cap.)" 

ff p (true) (true L. P. Cap.) 

~ # p (apparent) " (apparent L. P. Cap.)' 

true I.H.P. true m.e.p. 

~~ hypothetical I.H.P. ~ hypothetical m.e.p.' 

Then 

true work per cu.ft. gas, supplied 
hypothetical work per cu.ft. gas, supplied 

I.H.P. 



true -7V- 



true I.H.P. per cu.ft. gas supplied = (L. P. Cap.) .^^ 

"hypothetical I.H.P. per cu.ft. supplied" - ptif>fll I-H.P. 

(L.P.Cap.) 

_ true m.e.p. __.__ true L. P. Cap. _es 
"hypothetical m.e.p. ' hypothetical L. P. Cap. e\ 

This ratio can be used to convert from hypothetical work per cubic foo 
gas supplied to probable true work per cubic foot. 

Multi-stage Compressors are subject in each stage to all of the characteristic 
described for single stage to a greater or less extent. Valve resistance, frictio) 
and inertia affect the intake and discharge lines; heat transfer and leakag 
influence the form of compression and reexpansion lines, and the true capacit 
of the cylinder is made different from the apparent due to leakage, pressur 
and temperatures changes. 

In addition to these points it is useful to note one special way in which th 
multi-stage compressor differs from the single stage. The discharge of the fir? 
stage is not delivered to a reservoir in which the pressure is constant, but 
receiver of limited capacity. The average rate at which gas is delivered t 
the receiver must equal the average rate at which it passes to the next cylinde 
The momentary rate of supply and removal is not constantly the same, howeve 



WORK OF COMPRESSORS 



159 



''his causes a rise or fall of pressure. It is evident that this pressure fluctuation 
3 greatest for a small receiver. Very small receivers are not, however, used 
n gas compressors due to the necessity of cooling the gas as it passes from one 
tage to the next. To accomplish this a large amount of cooling surface must 
e exposed, requiring a large chamber in which it can be done. Thus, it is 
een that the hypothetical cycles assumed for multi-stage compressors do not 
ruly represent the actual cycle, but the difference can never be very great, 
ue to the large size of receiver which must always be used. 

Another way in which the performance of this multi-stage compressor 
ommonly differs from assumptions made in the foregoing discussions is in 



Del. Pr. 





\c 




EG. 38. Effect of Loss of Intercooling in Two-stage Compressors on Receiver-pressure and 
Work Distribution in the Two Cylinders. 

to intercooling. It seldom occurs that the gases enter all stages at 
F ie same temperature. In the several stages the temperature of the gases 
jjll depend on the amount of compression, on the cooling surface and on the 
tount and temperature of cooling water. The effect of variations in tem- 
irature upon the work and receiver pressures will be taken up later. It may 
noted now, however, that if all cooling water is shut off, the gas passes from 
ie cylinder to the next without cooling, there is no decrease in volume in the 
leiver. For simplicity take the case of zero clearance, two-stage (Fig. 38). 
ABCDEF be the cycle for perfect intercooling. AB and KD are the low- 
fressure capacities of the first and second stages respectively. If now, inter- 
ooling ceases, the gas will no longer change volume in the receiver. The 
iver gas, in order to be made sufficiently dense to occupy the same 



160 ENGINEERING THERMODYNAMICS 

volume (KD) as it did before, must be subjected to a greater pressure in th< 
first stage. The new receiver^ line will be K'D'. The work of the firs 
stage will therefore be ~ABWK'\ of the second stage K'D'GF, and the tota 
work in the new condition is greater than when intercoooling was perfect by ai 
amount represented by the area DCGE. 

In the case where clearance is considered, the effect is the same, except tha 
the increasing receiver pressure, increasing the ratio of compression of the firs 
stage, causes the volumetric efficiency of the first stage to become less, am 
hence lessens the capacity of the compressor. The effect on work per unit o 
capacity is the same as without clearance. 

The question as to how many stages should be used for a given compresso 
is dependent upon the ratio of compression largely, and so is due, first, to con 
siderations of economy, which can be understood from the foregoing sections 
second, for mechanical reasons, to avoid high pressures in large cylinders; third 
for thermal reasons, to avoid such high temperatures that the lubrication o 
the cylinders would be made difficult, or other dangers, such as explosions 
involved. 

Practice varies very widely as to the limiting pressures for single, tw< 
three or four-stage compressors. Air compressors of a single stage are con 
monly used for ratios of compression as high as 6 or 7 (75 to 90 Ibs. gage). Fc 
ratios greater than these, two-stage compressors are used, especially for larg 
sizes, up to ratios of 34 to 51 (500 to 750 Ibs. gage). Some three-stage con 
pressors are used for ratios as low as 11 or 14 (150 or 200 Ibs. gage), althoug 
installations of this nature are rare, and are warranted only when power is cost] 
and the installation permanent and continuously used to warrant the hig 
investment cost. As a minimum ratio for three stages, 11 (150 Ibs. gag* 
is used for large units, while a few small units compress as high as 135 or eve 
170 atmospheres (2000 or 2500 Ibs. gage). A notable use for the four-staj 
compressor is for charging the air flask of automobile torpedoes used by tl 
various navies, which use pressures from 1600 to 3000 Ibs. per square inci 
(110 to 200 atmospheres). These require special design of valves, cylindei 
and packings to withstand the extremely high pressures, small clearances, ai 
special precautions against leakage, due to the great loss of volumetric efficien 
and economy that would otherwise result. 

20. Work at Partial Capacity in Compressors of Variable Capacity, 
is seldom that a gas compressor is run continuously at its full capacity. 
the duty of the compressor is to charge storage tanks, it may be made to 
at its full capacity until the process is completed and then may be stopp 
entirely, by hand. Even where the compressed gas is being used continuou 
it is common practice to have a storage reservoir into which the compressor m 
deliver. This enables the compressor to deliver a little faster or slower tr 
the demand for a short period without a great fluctuation pressure in 
reservoir. For many purposes hand regulation is not sufficient or is 
expensive, hence the demand for automatic systems of capacity regulati 
These systems may be classified in a general way in accordance with the met! 



WORK OF COMPRESSORS 161 

)f driving. Some methods of power application permit of speed variations while 
>thers require constant speed. The former provides in itself a means of regulat- 
ng capacity within certain limits, while, if the compressor must run at constant 
ipeed, some additional means of gas capacity control must be provided. 

Compressors driven by an independent steam engine, or steam cylinders 
instituting part of the same machine may be made to run at any speed required 
jvdthin a very wide range and still kept low enough for safety. If driven 

gear, belt, rope, chain or direct drive from a source of power whose speed 
s constant, the speed of the compressor cannot be varied. Electric motor, 
5as-engine, oil-engine or water-power drives are subject to only limited speed 
ilteration and may, therefore, be placed in the constant speed class. 

Regulation of Capacity by Means of Speed Change. If the speed of the com- 
pressor is decreased below normal: 

1. Displacement of piston is decreased in proportion to the speed. 

2. Mean effective pressure, as to hypothetical considerations, is the same, 
due to the decrease of velocities in gas passages, the frictional fall of pres- 

jure during inlet and delivery is not so great, and hence the mean effective pres- 
|ire is not quite so great. If the compressor is multi-stage, since a smaller 
quantity of gas is passing through the intercooler, it is probable that the inter- 
jooling is more nearly perfect, thus decreasing the mean effective pressures in 
?jbhe succeeding stages. 

3. The volumetric efficiency is changed, due first to the fact that leakage 
s about the same in total amount per minute as at full speed, but the total 
juantity of gas being less, leakage is a larger percentage of the total; second, 
,he inertia of gases in the supply pipe, as well as their friction, has been decreased. 
The former tends to decrease vulmetric efficiency, while the latter may tend 
,o increase or decrease it. It may be expected that the true volumetric efficiency 

be somewhat greater at fractional speed than at full speed. 

For any compressor there is a speed of maximum economy above and below 
vhich the economy is less, though it may be that this most economical speed is 
greater than any speed of actual operation. 

It is not desirable at this point to discuss the effect of speed variation upon 
(he economy of the engine or other motor supplying the power. The reasoning 
foove applies to the term economy as applied to the compression effect obtained 
jer unit of power applied in the compression cylinder. It might be noted here, 
lowever, that the decrease of speed has little effect upon the mechanical efficiency 
p the compressor as a machine, since frictional resistance between solid parts 
ip|temains nearly constant, and, therefore, power expended in friction will vary 
(s the speed, as does approximately also the power to drive the compressor. 
The ratio of frictional power to total may then be expected to remain nearly 
Constant. 

Regulation of Capacity at Constant Speed may be accomplished in a number 

of ways : 

1. Intermittent running; 

2. Throttling the supply to compressor; 



162 ENGINEERING THERMODYNAMICS 

3. Periodically holding open or shut the intake valve; 

4. Closing intake valve before end of intake stroke, or holding intake valve 

open until compression stroke has been partially completed; 

5. Large clearance; 

6. Variable clearance, 

The first necessitates some means for stopping and starting the compressor, 
which is simple with electric drive, and may be accomplished in other cases by 
means of a detaching clutch or other mechanical device. The pressure in the 
reservoir is made to control this stopping and starting device by means of a 
regulator. This arrangement is made to keep the pressure in the reservoir 
between certain fixed limits, but does not maintain a constant pressure. The 
economy of compression in this case is evidently the same as at full speed 
continuous running, provided there is no loss in the driving system due to 
starting and stopping, which may not be the case. This method of 
regulation is used mainly for small compressors in which inertia is not 
great, such as supply the air brakes on trolley cars. The sudden change 
of load on the driving machinery would be too great if large compressors 
were arranged in this way. 

If the compressor whose capacity is regulated by intermittent running is 
multiti-stage, the constant supply of water to the intercoolers while the compres- 
sor is stopped will lower the temperature of the cooling surface, causing more 
nearly perfect intercooling when the compressor is started. Leakage, on the 
other hand, will permit the loss of pressure to a greater or less extent in the 
receivers while the compressor is stationary, which must be replaced after 
starting before effective delivery is obtained. 

Throttling the gas supply to the compressor has certain effects that may 
be studied by referring to Fig. 39, which represents the hypothetical cycles most 
nearly approaching this case. In order to reduce from the full-load low-pres- 
sure capacity, AB, to a smaller capacity, AE, the supply pressure is decreased 
by throttling to the pressure of B', such that B' and E lie on the same adiabatic 
The work area A'B'EA is entirely used up in overcoming the throttle resistance 
and is useless friction, so that economy is seriously reduced by this methoc 
of regulation. Such compressors may use almost as much power at partia 
as at full capacity. 

It is easily seen that this method of regulation would be undesirable, it 
only advantage being simplicity. 

The effect of throttling upon a multi-stage compressor may be illustrated a; 
in Fig. 40, by considering the two-stage compressor cycle without clearance 
ABCDEF. The ratio of compression of the first cylinder is determined wit! 

perfect intercooling by the ratio of displacements P c = P b (~\ . When th> 

supply pressure is throttled down to P 6 ', the new receiver pressure will b 
<Di\ 
-jj- } , a pressure much lower than P c . Hence the receiver pressur 

is decreased, less work done in the first stage, and far more than half the wor 



WORK OF COMPRESSORS 



163 



compression done in the second stage. If best-receiver pressure existed at 
normal capacity, it does not exist in the throttled condition. 

The intake valve may be held wide open or completely closed during one 
or more revolutions, thereby avoiding the delivery of any gas during that period. 
If the intake valve is held wide open, the indicator card would be as shown in 
Fig. 41A in full lines, ABCD, the dotted lines showing the cycle performed when 



c' 

i 
i 



A' 



-L.P.Cap Throttled > 



L.P.Cap at Full Load 



IG. 39. Effect of Throttling the Suction of One-stage Compressors, on Capacity and 

Economy. 



lormal operation is permitted. With the inlet valve open in this way there is 
ib loss of power due to friction of the gas in passage during both strokes, measured 
by the area within the loop. 

Closing the inlet valve and holding it shut will give an indicator card of the 
form EFG, Fig. 41 B, which will be a single line retraced in both directions 
except for probable leakage effects. If leakage is small, there will be but little 



164 



ENGINEERING THERMODYNAMICS 



area enclosed between the lines. At a high speed this might be expected 
incur less lost power than the former plan. 

Certain types of compressors are made with an intake valve controlled 
by a drop cut-off, much like the steam valve of the Corliss engine. The effect 
of this is to cut off the supply of gas before the end of the stroke, after which 
time the gas must expand hypothetically according to the adiabatic law. The 
return stroke causes it to compress along the same line continued up to the 
delivery pressure, as indicated by the line PEG, Fig. 4 1C. There is little work 



J 



1 



FIG. 40. Effect of Throttling Multi-stage Compressors on Receiver-pressure, Work Distri- 
bution, Capacity and Economy, 



lost in the process, none, if the line is superimposed as in the figure, and hence 
the process is the same as if only the cycle AEGD were performed. 

The same quantity of gas might have been entrapped in the cylinder by 
holding the intake valve open until the end of the stroke and on the return till 
the point E, Fig. 41D, was reached, then closing it. The same compression 
line EG will be produced. The line AB will not coincide with BE, due to 
friction of the gas in passages, and hence will enclose between them a small 
area representing lost work, which may be no larger than that lost in the process 
EFE. 



WORK OF COMPRESSORS 



165 



If such an automatic cut-off were applied only to the first stage of a multi- 
stage compressor, the effect would be to lower receiver pressures as in the throt- 
tling process. To avoid this, the best practice is to have a similar cut-off to act 
on the supply to all of the stages. If this is properly adjusted, the receiver 
pressures can be maintained the same as at full load. An additional advantage 
of this system is that even if the compressor is to be used for a delivery pres- 
sure for which it was not originally designed, the relative cut-offs may be so 
adjusted as to give and maintain best-receiver pressure. 





1 -* 


l 
















\ 
\ 
\ 














1 
I 
\ 


\ 


v 
\ 












\ 
\ 
\ 




\ 

N > 


x 










\ 

\ 






D 


" > *^, 


-,^_ 


(Su 


x Pr. 


A 


V 




B 








C 



B 



(A) 



D G 






\ 






\ 



(B) 



C 



D 



FIG. 41. Control of Compressor Capacity by A. Open Inlet Valve; B. Closed Inlet Valve ; 
C. Suction Cut-off; D. Delayed Suction Closure. 

Since the low-pressure capacity per cycle of a compressor involves clearance 
and ratio of compression as two of its variables, it is possible to change capacity 
by changing either the clearance or the ratio of compression. fc 



(L. P. 



(227) 



Assuming that clearance is a fixed amount and not zero, it is evident that an 
increase in the ratio of compression decreases the capacity, and when it has 



166 



ENGINEERING THERMODYNAMICS 



reached a certain quantity will make the capacity zero. 

i 



If the clearance is 



large, making the coefficient of R p s large in the equation the effect of a change 
in that factor is increased. Fig. 42 indicates the hypothetical performance 
of a compressor with large clearance. When the pressure of delivery is low 
(say P e ) the capacity is large, AB. The cycle is then A BCD. An increase 
of the delivery pressure to P c f changes the cycle to A'BC'D' and the low pres- 
sure capacity is A'B. If the compressor is delivering to a receiver from which 
no gas is being drawn, the delivery pressure will continue to rise and the capacity 




FIG. 42. Variation of Compressor Capacity with Rise of Delivery Pressure, Fixed Clearance, 

Pressure for Zero Delivery. 

to decrease till the capacity approaches zero as the delivery pressure approaches 
the pressure P e as a limit. 

(limiting del.pr.) =(sup.pr.)(-~ j ..... (228) 

:. \ C I 

When the limiting condition has been reached and the capacity has become 
zero, the compression and reexpansion lines coincide and enclose zero area 
between them; hence, the mean effective pressure and the indicated horse- 
power are zero," for the hypothetical case. Leakage will prevent a perfect 
coincidence of the lines and cause some power to be required in addition to that 
of friction. 

Such a simple method of regulation as this is used for some small com- 
pressors driven constantly from some source of power used primarily for 
other purposes. When it is not necessary to have a constant delivery pressure, 



WORK OF COMPRESSORS 



167 



J but only to keep it between certain limits, this may be made use of, especially 
if the limits of pressure are quite wide. 

The expression for low-pressure capacity Eq. (227), suggests the possi- 
bility of decreasing capacity by the increase of clearance. The effect of this is 
shown in Fig. 43. The original compression cycle (full capacity) is shown by 
ABCD, with a clearance volume of cD, so that the axis of zero volume is OP. 
Increasing the clearance to c'D causes a smaller volume C'D to be delivered 
and due to the more sloping re-expansion DA', a smaller volume of gas is 
taken in, A 'B. 

It has been shown in previous sections that clearance has no effect upon the 
economy of a compressor so far as hypothetical considerations are regarded. 
In practice it is found that a slight loss of economy is suffered at light load, 



o' 



P D 



C' C 









\ 



-L.P. Cap Full Load- 
D- 



L.P. Cap Part Load- 



FIG. 43. Variation of Compressor Capacity by Changing Clearance. 

as might be expected, due to greater leakage per unit of capacity. The addi- 
tional clearance is provided in the form of two or more chambers connected to the 
clearance space of the compressor by a passage in which is a valve automatically 
controlled by the receiver pressure. 

In the multi-stage compressor, decreasing the capacity of the first stage by 
an increase of its clearance would evidently permit a decrease of receiver pres- 
sures unless the capacity of each of the various stages is decreased in the same 
proportion. Eq. (132) gives the condition which must be fulfilled to give best 
receiver pressure for a two-stage compressor. 



J_1 



2* 



168 



ENGINEERING THEEMODYNAMICS 



Since Di, Z>2, and R p remain fixed, for any chosen value of clearance of the 
first stage, ci, the clearance of the second stage, C2, to give best-receiver 
pressure can be found, 



C 2 = 



(Rp2~s-l)R p 2sD 2 



(229) 



For every value of first-stage clearance there is a corresponding clearance of 
second ^stage that will give best-receiver pressure, found by this equation. Sim- 
ilar reasoning can be applied to three- or four-stage compressors. 

2i. Graphic Solution of Compressor Problems. In order to obviate the 
necessity of working out the formulas given in this chapter each time a prob- 
lem is to be solved, several of them have been worked out for one or more 
cases and results arranged to give a series of answers graphically. By the 
use of the charts made up of these curves many problems may be solved 
directly and in many others certain steps may be shortened. A description of 
each chart, its derivation and use is given in subsequent paragraphs. 

Chart, Fig. 44. This chart gives the work required to compress and deliver 
a cubic foot of (sup.pr.) air or the horse-power to compress and deliver 1000 cu.ft. 
of (sup.pr.) air per minute if the ratio of pressure (del.pr.)^ (sup.pr.), the value 
of s and the (sup.pr.) are known, and compression occurs in one stage. The 
work or H.P. for any number of cubic feet is directly proportional to number 
of feet. The curves are dependent upon the formulas, Eq. (31), for the case 
when 8 = 1, and Eq. (51) for the case when s is not equal to 1. These formulas 
are: 

Eq. (31), W per cu.ft. = 144 (sup.pr) log e R p ', 



(51), W per cu.ft. = 144 - (sup.pr.) 
sl 



These equations are difficult to solve if an attempt is made to get a relation 
between the work and ratio of pressures. This relation may, however, be 
worked out for a number of values of pressure ratios and results plotted to 
form a curve by which the relation may be had for any other ratio within 
limits. This has been done in this figure in the following manner: 

On a horizontal base various values of R p are laid off, starting with the value 
2 at the origin. The values for work were then found for a number of values 
of R p with a constant value of (sup.pr.) and s. A vertical work scale was 
then laid off from origin of R p and a curve drawn through the points found 
by the intersection of horizontal lines through values of work, with vertical 
lines through corresponding values of R p . The process was then repeated for 
other values of s and curves similar to the first, drawn for the other values 
of s. From the construction so far completed it is possible to find the work per 
cubic foot for any pressure ratio and any value of s for one (sup.pr.) by pro- 



WORK OF COMPRESSORS 



169 




UT -bs aad 'sqi ('JcT 'dns) 



17 o ENGINEERING THERMODYNAMICS 



WORK OF COMPRESSORS 



171 



CO *. -1 




o 



I 






I! 



o 

I 
o 

I 



172 



ENGINEERING THEKMODYNAMICS 



I . . . 




SJ 3$ 8 

sqy *ui -nbs aad ajy ('Jd 'dn 



WORK OF COMPRESSORS 173 

C er to (sup.pr.) = 15 and down, and the horse-power will be found to be 13.6 
4 before by use of formulas. 

Chart, Fig. 47. This chart is for finding the (m.e.p.) of compressors. In the 

c?e of multi-stage compressors with best-receiver pressure and perfect inter- 

pling, the (m.e.p.) of each cylinder may be found by considering each cylinder 

4 a single-stage compressor, or the (m.e.p.) of the compressor referred to the 

P. cylinder may be found. 

The chart depends on the fact that the work per cubic foot of (sup.pr.) gas 

equal to the (m.e.p.) for the no clearance case and that the (m.e.p.) with 

arance is equal to the (m.e.p.) for no clearance, times the volumetric 

iciency. Diagrams 1, 2 and 4 are reproductions of Figs. 45, 46 and 47 to a 

aller scale and hence need no explanation as to derivation. Their use may 

briefly given. From the proper ratio of pressures project upward to the 

Dper curve, then horizontally to the (sup.pr.) and downward to read work per 

bic feet of (sup.pr.) gas. 

The volumetric efficiency diagram was drawn in the following manner :- 

om Eq. (65) vol. eff. = (l+c cR p s ), showing that it depends upon three 

riables, R P , c and s. A horizontal scale of values of. R p was laid off. Values 
i 

R p s were found and a vertical scale of this quantity laid off from the same 
gin as the R P values. Through the intersection of the verticals from various 
lues of R p with the horizontals drawn through the corresponding values of 

P) S for a known value of s, a curve of this value of s was drawn. In a similar 

y curves of other values of s were drawn. From the construction so far 

i 

aipleted it is possible to find the value of (R p ) 



174 



ENGINEERING THERMODYNAMICS 




f 1 8 9 10 11 12 13 14 

tfatio of Pressures 

81 77 70 63 5fe 49 4\i 35 28 21 14 7 

Work per Cu. Ft. of (Sup. Pr.) Gas-H44 



FIG. 



f f 
'7 70 63 

Work 

47. Mean Effective Pressure of Compressors, One-, Two-, and Three-stages. 



WORK OF COMPRESSORS 
Initial Pressure Lbs. per Sq. In. Abs. 



175 




11 12 



14 15 



7 8 9 10 

Ratio of Pressures 

70 63 56 49 42 35 28 21 14 7 

AVork per Cu. Ft. of (Sup. Pr.) Gas-M44 



FIG. 47. Mean Effective Pressure of Compressors, One-, Two-, and Three-stages. 



176 ENGINEERING THERMODYNAMICS 

each to be a single-stage compressor and remembering that (1 rec.pr.) becomes 
(sup.pr.) for second stage, and (del.pr.) for first stage and that (2 rec.pr.) 
becomes (sup.pr.) for third stage, (del.pr.) for second stage. The (m.e.p.) reduced 
to low-pressure cylinder is found by taking work per cubic feet of (sup.pr.) 
gas and multiplying by volumetric efficiency of low-pressure cylinder. 

To illustrate the use of this curve the example of Section (16) may be 
solved. Projecting upward from the pressure ratio of 9.35 to the line of s = lA 
and then over to (sup.pr.) = 15 in diagram 4, since compression is three stage 
and from 15 Ibs. per square inch to 140 Ibs. per square inch, work per cubic 
foot or (m.e.p.), is found for no clearance to be 37.8 abs. per square inch. Since 
best-receiver pressure assumed is 31.6, which gives a ratio of 2.1 for the low- 
pressure cylinder. From diagram 3, by projecting upward from R P = 2.I and 
over to the 5 per cent clearance line volumetric efficiency is 96.5. The product 

(m.e.p.)Lan 
gives (m.e.p.)reduced to low-pressure cylinder and is 36.5. From the 

formula, horse-power is found to be 358 as before. 

Chart, Fig. 48. As mentioned in Section 18, there is one (sup.pr.), which 
for a given (del.pr.) will give the maximum work of compression. The chart, 
Fig. 48, originated by Mr. T. M. Gunn, gives a graphical means of finding this 
value of (sup.pr.) when the (del.pr.), clearance and value of s are known. It also 
gives on the right-hand of the chart a means for finding the (m.e.p.) for this 
condition. The figure was drawn by means of Eqs. (214) and (218). For the 
value of s = l the ratio of (del.pr.) to (sup.pr.) was found for cases of clearances 
from per cent to 15 by means of Eq. (218) by trying values of this ratio which 

/del.pr. \ c /del.pr. \ 

would fulfill the condition of the equation, logel p ) = ~jqL~( gup pr ). 

For values of s not 1, Eq. (214) was used, and a set of values of R p found 
for the values of s = 1.4 and 1.2 by trial, the correct value of R p being that which 
satisfied the equation, 



/deLpr 
Vsup. 



pr\ = __ [" __ s-1 /del.pr. \ .1 
.pr./ 1+c L ~ * \suP-pr. / J 



As an example the work for the case where s = 1.2 and c = 10 per cent is given. 
Try fl, = 2.6, then, R^ = 



= 1.091(1. 1-.01667X2.218), 
= 1.161 



WORK OF COMPRESSORS 



\ 



8 # 8 $ 

(d*a*tn) jo OT:VBH mmnproH* 




'saanssaajjo ORBH 



178 



ENGINEERING THERMODYNAMICS 



R p = (1.161) 6 = 2.45, which shows the value of 2.6 to be incorrect. For a second 
trial take 2.45, and then, 

,*, = 1.091(1.1 - .01667 X2.45' 833 ), 
= 1.1627 

= (1.1617)6 = 2.458, 

which is sufficiently close. Therefore the value of R p for s = 1.2 and clearance = 
10 per cent, is 2.45; that is, maximum work will occur for a given (del.pr.) 

when (sup.pr.) is ^5 times the ( del -P r -) 

When the values for R P had been obtained a horizontal axis of values of s 
and a vertical one of R P values, were laid off and the points for clearance curves 
laid off to their proper values referred to these axes. Through points as plotted 
the clearance lines were drawn. The right-hand diagram was plotted in a 
similar manner from Eqs. (213) and (217), for s not equal to 1 and equal to 
1 respectively. 

The latter formula was rearranged in the form 



(del.pr. / " R p * E " 

the last term being found from curve of Fig. 45. The value of R p for each 
value of the clearance was taken from the left-hand diagram, and substituted 

in the above expression to obtain ( -r-p L Jfor the case of s = 1. Eq. (213) was 

put in form 

(m.e.p. \ _ * 
deLpr./ (*- 



and values of R p for each value of trie clearance found in the left-hand diagram 
were substituted, together with E v values from Fig. 45 and the value of ( , ', ' j 

found for each case of clearance when s = 1.4. When the points for s = l and 
s = 1.4 had been found, a horizontal axis of values of s and a vertical one of 
values of R p were laid off, and points for the clearance curves plotted as for 
the left-hand diagram and the curves drawn in. 

To find the (sup.pr.) to give maximum work for any (del.pr.) it is only 
necessary to project from the proper value of s to the proper clearance curve, 
and then horizontally to read the value of R p . The (del.pr.) divided by this 
gives the (sup.pr.) desired. To obtain the (m.e.p.) project upward from the 

value of s to the clearance curve, then horizontally to read the ratio ( -rr- I 

\ del.pr./ 

The (del.pr.) times this quantity gives the m.e.p. 

As an example of the use of this chart let it be required to find the (sup.pr.) 
for the case of maximum work for 9X12 in. double-acting compressor running 
200 R.P.M., having 5 per cent clearance and delivering against 45 Ibs. per square 
inch gage. Also the horse-power. Compression such that s = 1.3. 



WORK OF COMPRESSORS 



179 



Projecting from the value 1.3 for s on the left-hand diagram to the line of 
5 per cent clearance find R p to be 2.8, hence (sup.pr.) = 2^ = 21.4 Ibs. per square 
inch absolute = 6.4 Ibs. per square inch gage. Again, projecting from value 1.3 

for s on right-hand diagram to line of 5 per cent clearance find that;', 6 '*; = .383, 

(del.pr.) 



hence (m.e.p). = 23 and 



I.H.P. 



23X1X64X400 



= 17.8. 



1.00 



IM 

II 



I 



ft .80 

I 

o .78 



.7-1 



.72 






S=l.l 



S = 



8 = 1.3 



8 = 1.4 
8 = 1.3 



8 = 1.5 



8=1.4 



8 = 1.5 



2 3 4 5 67 8 9 10 11 12 13 14 15 

Ratio of Pressure = Rp 

Note; Solid Lines = 3 Stage; Broken Lines = 2 Stage 

FIG. 49. Relative Work of Two- and Three-stage Compressors Compared to Single Stage. 



180 ENGINEERING THERMODYNAMICS 

The chart was made by laying off on a horizontal base a scale of pressure 
ratios. From the same origin a scale of work for two or three stage divided by 
the work of one stage was drawn vertically. For a number of values of R P 
the work to compress a cubic foot of gas was found for one, two and three stage 
for each value of s. The values found by dividing the work of two or three 
stage by the work of single stage were plotted above the proper R p values and 
opposite the proper ratio values and curves drawn through all points for one 
value of s. To find the saving by compressing in two or three stages project 
from the proper R p value to the chosen s curve for the,desired number of stages, 
then horizontally to read the ratio of multi-stage to one-stage work. This value 
gives per cent power needed for one stage that will be required to compress 
the same gas multi-stage. Saving by multi-stage as a percentage of single 
stage is one minus the value read. 

To illustrate the use of this chart, find the per cent of work needed to 
compress a cubic foot of air adiabatically from 1 to 8J atmospheres in two 
stages Compared to doing it in one stage. From examples under chart Nos. 
44 and 46 it was found that work was 6300 ft.-lbs. and 5320 ft.-lbs. respec- 
tively, for one- and two-stage compression, or that two stage was 84.5 per cent 
of one stage. From R p , 8J project up on Fig. 49 to s = 1.406 for two stage and 
over to read 84.6 per cent, which is nearly the same. 

Chart, Fig. 50. This chart, designed by Mr. T. M. Gunn, shows the 
economy compared to isothermal compression. 

The chart was drawn on the basis of the following equation: 

Economy (isothermal) = g^.J!thermaJ (no clearance) 

m.e.p. actual -r-E v actual 

(sup.pr.) 



s-l 



-^r (sup.pr.) (R p s -1) 

o J. 

\OgeRp 



Values of this expression were worked out for each exponent, for assumed values 
of R p . A scale of values of R p was laid off horizontally and from the same 
origin a vertical scale of values of the ratio of isothermal to adiabatic. The 
results found were then plotted, each point above its proper R p and opposite its 
ratio value. Curves were then drawn through all the points found for the 
same value of s. In a similar way a set of curves for two stage and a set for 
three stage were drawn. 

This chart is also useful in obtaining the (m.e.p.) of the cycle if the (sup.pr.) 
and the volumetric efficiency of the cylinder be known. A second horizontal 
scale laid off above the R p scale shows the (m.e.p.) per pound of (sup.pr. for) 
the isothermal no-clearance cycle. This is found to be equal to log e R P} since 






WORK OF COMPRESSORS 



r . 




qgiiAv pamluioo ^uiouoo^ 



182 



ENGINEERING THERMODYNAMICS 



the (m.e.p.) for no clearance is equal to the work per cubic foot of (sup.pr.) gas, 
which, in turn, for the isothermal case is (sup.pr.) log* R p or loge R v when 

(sup.pr.) = 1. 

Knowing the ratio of pressures, economy compared to isothermal can be 
found as explained above. Also knowing R p the (m.e.p.) per pound initial is 
found from the upper scale. 

Since the latter quantity is assumed to be known, by multiplying it by 
factor just found there is obtained (m.e.p.) isothermal. Since volumetric efficiency 
is assumed known, all the factors are known for the first equation given above 
which, rearranged, reads 



. (m.e.p.) actual = 



m.e.p. isothermal (no clearance) 
(economy isothermal) -f- E v 



Chart, Fig. 51. This chart is drawn to give the cylinder displacement for a 
desired capacity, with various values of R p , s and clearance. From the formula 
Eq. (64) 



(L. P. Cap.)=Z>(l+c-c# p *). 
The right-hand portion of the diagram is for the purpose of finding values 

of (Rp) * for various values of R p and s, and is constructed as was the similar 
curve in Fig. 45. The values of the lower scale on the left-hand diagram give 

JL 
values of D = (L. P. Cap.)-J-(l+c cR p s ), where capacity is taken at 100 cu.ft., 

this scale was laid out and the clearance curves points found by solving the 

j_ 

above equation for various values of (R p ) s for each value of c. To obtain the 
displacement necessary for a certain capacity with a given value of R PJ c and 
s, project upward from R p to the proper s curve across to the c curve and down to 
read displacement per hundred cubic feet. Also on the left-hand diagram are 
drawn lines of piston speed, and on left-hand edge a scale of cylinder areas 
and diameters to give displacements found on horizontal scale. To obtain 
cylinder areas or approximate diameters in inches project from displacement to 
piston speed line and across to read cylinder area or diameter. Figures given 
are for 100 cu.ft. per minute. For any other volume the displacement and 
area of cylinder will be as desired volume to 100 and diameters will be as 
Vdesired volume to 100. 

As an example, let it be required to find the low-pressure cylinder size for a 
compressor to handle 1500 cu.ft. of free air per minute. Receiver pressure to 
be 45 Ibs. per square inch gage and (sup.pr.) to be atmosphere. Piston speed 
limited to 500 ft. per minute. Compression to be so that s = 1.4 and clear- 
ance =4 per cent. Projecting upward from ft p = 4 tos = 1.4, across to c = 4%, 
and down to piston speed = 500, find the diameter of a cylinder for 100 cu.ft. 

per minute is 6.3. For 1500 cu.ft. diameter will be as Vl5X6.3 = 3.9X6.3 = 
ins. 



WORK OF COMPRESSORS 



183 




8 4 ENGINEERING THERMODYNAMICS 



GENERAL PROBLEMS ON CHAPTER II. 

Prob. 1. One hundred cubic feet of H 2 S are compressed from 15 Ibs. per square 
ich absolute to 160 Ibs. per square inch absolute. 

(a) Find work done if compression occurs isothermally in a no-clearance one-stage 
compressor; 

(6) Adiabatically in a two-stage, no-clearance compressor; 

(c) Adiabatically in two-stage compressor each cylinder having 5 per cent clearance ; 

Prob. 2. Air is being compressed in three plants. One is single-stage, the second is 
sro-stage, and the third is three-stage Considering the compressors to have no clear- 
ace and 1000 cu.ft. of free air compressed adiabatically per minute from atmosphere 
3 150 Ibs. per square inch gage, what will be the horse-power required and cylinder 
Lzes in each case? 

Prob. 3- A two-stage compressor with 5 per cent clearance n the h'gh and 3 per 
ent in the low-pressure cylinder is compressing air from 14 Ibs. per square inch ab- 
olute to 125 Ibs. per square inch gage. What is the best-receiver pressure and what 
lust be the size of the cylinders to handle 500 cu.ft. of free air per minute? 

Prob. 4. A manufacturer gives a capacity of 970 cu.ft. for a 22 J X24 in. single- 
tage compressor running at 142 R.P.M. when working pressures are 50 to 100 Ibs. per 
quare inch gage. What would be the clearance for each of these pressures assuming 
= 1.4? 

Prob. 5. The card taken from a single-stage compressor cylinder showed an appar- 
nt volmetric efficiency of a 95 per cent and a mean value for s of 1.3. What is the 
learance and what would be the (m.e.p.) for the ratio of pressures of 6? 

Prob. 6. A compressor with double-acting cylinder 12x14 ins., having 6 per cent 
learance, is forcing air into a tank. Taking the volumetric efficiency as the mean of 
hat at the start and the end, how long w 11 it take to build up 100 Ibs. per square inch 
;age pressure in a tank of 1000 cu.ft. capacity if the speed is 100 M.R.P. and compres 
ion is isothermal, and how long wi 1 it take to do it if the compression is adiabatic and 
he ah* in the tank does not cool during filling? What is the maximum attainable pressure? 

Prob. 7. It is desired to supply 1000 cu.ft. of air per minute at a pressure of 80 Ibs. 
>er square inch gage. A two-stage compressor is to be used, the clearance in low pres- 
ure of which is 3 per cent. What must be the displacement of the low-pressure cylin- 
ler and what will be the horse-power of the compressor? 

Prob. 8. The low-pressure cylinder of a compressor is 18x24 ins. and has a clear- 
tnce of 4 per cent. The receiver pressure is 60 Ibs. per square inch absolute. The high- 
>ressure cylinder has a 5 per cent clearance. What must be its diameter, stroke being 
ame as low, so that compressor will operate at its designed receiver pressure? 

Prob. 9. The discharge pressure of a two-stage compressor is 120 Ibs. per square 
och absolute and the supply pressure is 15 Ibs. per square inch absolute The compressor 
s 10x16x12 ins. The clearance in the low-pressure cylinder is 3 per cent. What 
Qust be the clearance in the high-pressure cylinder for the machine to operate at best- 
eceiver pressure? Compression is adiabatic. 

Prob. 10. If the clearance in the high-pressure cylinder df Prob. 9 were reduced 
o 3 per cent, would the receiver pressure increase or decrease, how much and why? 

Prob. 11. If the discharge pressure in Prob. 9 fell to 100 Ibs. per square inch absolute, 
vhat would be the new best-receiver pressure and why? Would the original clearance 
illow the new best-receiver pressure to be maintained? 



PROBLEMS ON CHAPTER II 185 

Prob. 12. The discharge pressure for which a 20 Jx 32^x24 in. compressor is 
lesigned, is 100 Ibs. per square inch gage, supply pressure being 14 Ibs. per square inch 
Absolute. The d'scharge pressure is raised to 125 Ibs. per square inch gage. The 
ilearance on the high-pressure cylinder can be adjusted. To what value must it be 
;hanged to enable the compressor to carry the best-receiver pressure for the new 
lischarge pressure? Low-pressure clearance is 5 per cent at all times and corn- 
session being adiabatic. 

Prob. 13. A manufacturer builds his 15^X25^X18 in. compressors with low-pres- 
ure cylinders of larger diameter for high altitude work. What would be the diameter 
>f a special cylinder for this compressor to work at an altitude of 10,000 ft. and what 
v r ould be the horse-power per cubic foot of low-pressure air in each case? 

Prob. 14. A three-stage compressor has 4 per cent clearance in all the cylinders. The 
ow-pressure cylinder is 34x36 ins., delivery pressure 200 Ibs. per square inch gage, 
upply pressure 14 Ibs. per square inch absolute. What must be the size of the other 
Cylinders for the machine to operate at best-receiver pressure. 

Prob. 15. The cylinders of a two-stage compressor are given as 10J and 16^ ins., 
he stroke being 12 ins. The machine has a capacity of 440 cu.ft. per minute at 160 
I.P.M., the supply pressure is 14 Ibs. per square inch absolute and the delivery 
pressure 100 Ibs. per square inch gage. What is the clearance of each cylinder? 

Prob. 16. To perform the required useful refrigerating effect in an ice plant, it is 
iccessary to compress 250 cu.ft. of ammonia vapor per minute from 30 Ibs. per square 
ach gage to 150 Ibs. per square inch gage. What must be the size of the compressor 
o handle this at 75 R.P.M. if it be double acting and the clearance be 5 per cent? 

Prob. 17. The maximum delivery pressure of a type of compressor is controlled by 
aaking the clearance large so that the volumetric efficiency will decrease as the pressure 
ises and become zero at the desired pressure. What must be the clearance for a single- 
tage compressor where the supply pressure is 14 Ibs. per square inch absolute and the 
aaximum delivery pressure 140 Ibs. per square inch absolute? What will be the volu- 
aetric efficiency of the same machine at a delivery pressure of J the maximum? At |? 

Prob. 18. A three-stage compressor has a clearance of 5 per cent in each cylinder. 
Vhat must be the cylinder ratios for the best-receiver pressures when the machine is 
ompressing to 170 Ibs. per square inch gage from atmosphere? 

Prob. 19. Show why it was very essential to keep the clearance low in cylinders of 
hree-stage compressor used for compressing air for air-driven cars, where the delivery 
ressure carried was 2500 Ibs. per square inch, by assuming numerical data and 
alculating numerical proof. 

Prob. 20. With water falling 150 ft. and used to compress air directly, how many 
ubic feet of air could be compressed per cubic foot of water? 

Prob. 21. Air is compressed from atmosphere to 150 Ibs. per square inch absolute, 
lothermally in one stage. How much more work would be required per cubic foot if 
ompression were adiabatic? How much of this excess would be saved by compressing 
wo stage? Three stage? 

Prob. 22. 150 I.H.P. is delivered to tluTair cylinders of a 14| X22i Xl8 in. compres- 
)r, running at 120 R.P.M. The supply pressure is 15 Ibs. per square inch absolute, 
'he volumetric efficiency as found from the indicator card is 95 per cent. What was 
e discharge pressure? 

Prob. 23. The clearance in the high-pressure cylinder of a compressor is 5 per cent, 
hich allows the compressor to run with the best-receiver pressure for a discharge of 
00 Ibs. per square inch absolute when the compressor is at sea-level. What would the 



L86 ENGINEERING THERMODYNAMICS 

clearance be if the discharge pressure were kept the same and the altitude were 10,000 
ft. to keep the best-receiver pressure? 

Prob. 24. How many cubic feet of supply-pressure air may be compressed per minute 
from 1 to 8 atmospheres absolute by 100 horse-power if the compression in all cases is 
adiabatic? 

(6) Three stage, no clearance; 

(c) Two stage with 5 per cent clearance; 

(d) Single stage with 5 per cent clearance; 
(a) Two stage, no clearance 

Prob. 25. The capacity of a 14iX22ixl4 in. compressor when running at 140 
R.P.M. is said to be 940 cu.ft. for working pressures of 80 to 100 Ibs. per square inch 
gage and atmospheric supply at sea level. Check these figures. 

Prob. 26. What horse-power would be required by an 18ix30ix24 in. compressor; 
operating at 100 P.P.M. and on a working pressure of 100 Ibs. per square inch gage if; 
the clearance in low-pressure cylinder is 4 per cent? What would be the capacity? 

Prob. 27. By means of water jackets on a compressor cylinder the va ue for s of com-j 
pression curve in single-stage machine is lowered to 1.3. Compare the work to com] 
press 1000 cu.ft. of air from 1 to 8 atmospheres absolute with this condition with thaj 
required for isothermal and adiabatic compression. 

Prob. 28. What must be the size of cylinders in a three-stage compressor for com-l 
pressing gas from 50 Ibs. per square inch absolute to 600 Ibs. per square inch absolute 
when s equals 1.3 and each cylinder has 3 per cent clearance. Compressor to run ai| 
100 R.P.M. to be double acting and handle 1000 cu.ft. of gas per hour? 

Prob. 29. How many cu.ft. of free air could be compressed per minute with an avail- 
able horse-power of 1000 H.P. from atmosphere to 150 Ibs. per square inch gage; (a) 
if compression is isothermal; (6) if compression be single-stage adiabatic; (c) if com- 
pression be three-stage adiabatic? 

Prob. 30. A single-stage compressor is compressing air adiabatically at an altitude 
of 6000 ft. to a pressure of 80 Ibs. per square inch gage. The cylinder has 2 per cent 
clearance. What must be the s'.ze of the cylinder to compress 2000 cu.ft. of free air pel 
minute if the piston speed is limited to 600 ft. per minute? What, if the clearance be 
zero? 

Prob. 31. What would be the size of the two-stage compressor for same data as in 
Prob. 30? 



CHAPTER III 

VORK OF PISTON ENGINES. HORSE POWER AND CONSUMPTION OF 
PISTON ENGINES USING STEAM, COMPRESSED AIR, OR ANY OTHER 
GAS OR VAPOR UNDER PRESSURE. 

1. Action of Fluid in Single Cylinders. General Description of Structure 
ind Processes, The most commonly used class of engines is that in which 
,he operation is dependent on the pushing action of high-pressure fluids on 
)istons in cylinders, and this includes all piston steam engines of the recip- 
ocating or straight-line piston path group as well as the less common 
otary group, having pistons moving in curved and generally circular paths. 
Cn these same engines there may be used compressed air as . well as steam, 
md equally as well the vapors of other substances or any other gases, without 
shange of structure, except perhaps as to proportions, providing only that the 
substance to be used be drawn from a source of supply under high pressure, 

admitted to the cylinder, there used and from it discharged or exhausted 
;o a place of lower pressure. This place of lower pressure may be the open 
iir or a closed chamber; the used fluid may be thrown away and wasted or 
ised again for various purposes without in any way affecting the essential 
Drocess of obtaining work at the expense of high-pressure gases or vapors. 
[t is evident that, regarding a piston as a movable wall of a cylinder, when- 
3ver a fluid acts on one side with greater pressure than on the other, the 
Diston will move toward the lower pressure end of the cylinder, and in so 
noving can exert a definite force or overcome a definite resistance, measured 

the difference in pressure on the two sides and the areas exposed to the 
Dressure. It is not so evident, but just as true, that the piston may be made 
io move from one end of the cylinder to the other when the average pressure 

one side is greater than the average pressure on the other, and also do work 
3ven if the excess of pressure should reverse in direction during the stroke 
provided only some energy storage device is added. In the common 
steam or compressed-air engine this is a flywheel with the usual connecting 
"od and crank mechanism, uniting the reciprocating piston movement with 
ihe continuous rotary movement of the flywheel mass. In certain forms 
pumps the energy is stored in extra cylinders at times of excess and given 
Nit at times of deficiency in the path of the piston, so that its motion from 
to end of cylinder may not be interrupted even if the pressure on the 
driving side should fall below that on the resisting side, assuming, of course, 
he average pressure for the whole stroke to be greater on the driving side 
ban on the resisting side. 

187 



188 



ENGINEERING THERMODYNAMICS 



It appears, therefore, that piston movement in engines of the common 
form and structure, and the doing of work by that movement is not a 
question of maintaining a continuously greater pressure on one side than 
on the other. On the contrary, the process is to be studied by examination 
of the average pressure on the driving side and that on the resisting side, 
or by comparing the whole work done on one side with the whole work 
done on the other side by the fluid. The work done by the fluid on one 
side of a piston may be positive or negative, positive when the pressures are 
assisting motion, negative when they are resisting it. It is most con- 
venient to study the action of fluids in cylinders by considering the 
whole action on one side from the beginning of movement at one end to 
the end of movement at the same point, after the completion of one complete 
forward and one complete return stroke. All the work done by the pressure 
of the fluid on the forward stroke on the side of the piston that is apparently 
moving away from the fluid is positive work, all the work done by the pres- 
sure of the fluid on the same side of the piston during the return stroke is 
negative, and for this stroke the side of the piston under consideration is 
apparently moving toward the fluid or pushing it. 

For the complete cycle of piston movement covering the two strok 
the work done on one side is the algebraic sum of the forward strok' 
work, considered positive, and the back stroke work, considered negative. 
During the same time some pressures are acting on the other side of the 
piston, and for them also there will be a net work done equal to the cor- 
responding algebraic sum. The work available for use during the comple 
two strokes, or one revolution, will be the sum of the net work done by th 
fluid on the two sides of the piston during that time, or the algebraic sum o 
two positive and two negative quantities of work. Methods of analysis of 
work of compressed fluids in cylinders are consequently based on the action i 
one end of a cylinder, treated as if the other end did not exist. 

Just how the high-pressure fluid from a source of supply such as a boiler 
or an air compressor is introduced into one end of a cylinder, how it is treated 
after it gets there, and how expelled, will determine the nature of the varia- 
tion in pressure in that end that acts on that side of the piston, and th 
are subjects to be studied. To determine the work done in the cylinde 
end by the fluid, it is necessary to determine laws of pressure change with 
stroke, and these are fixed first by valve action controlling the distribution 
of the fluid with respect to the piston and second by the physical properties 
of the fluid in question. 

It is necessary that the cylinder be fitted with a valve for getting fluid 
into a cylinder, isolating the charge from the source of supply and getting i 
out again, and it may be that one valve will do, or that two or even more 
are desirable but this is a structural matter, knowledge of which is assumed 
here and not concerned with the effects under investigation. The first ste, 
in the process is, of course, admission of fluid from the source of supply to th 
cylinder at one end, which may continue for the whole, or be limited to a 



WOEK OF PISTON ENGINES 189 

3f the stroke. When admission ceases or supply is cut off before the end 

| 3f the stroke there will be in the cylinder an isolated mass of fluid which 

will, of course, expand as the piston proceeds to the end. Thus the forward 



190 



ENGINEERING THERMODYNAMICS 



may evaporate; the resistance through valves will always make the cylinder 
pressure during admission less than in the supply chamber and greater during 
exhaust than the atmosphere or than in exhaust receiver or condenser and may 
through the valve movements make what might have been a constant-pres- 
sure straight line become a curve. There will, by reason of these influences, 
encountered in real engines, be an almost infinite variety of indicator cards 
or pressure-volume cycles for such engines. 



\ 



A-B STEAM ADMITTED 
B-C " EXPANDED 
C-D EXHAUSTED 
D-A " COMPRESSED 




A' 



FIG. 52. Diagram to Indicate Position of Admission, Cut-off, Release, Compression on) 

Engine Indicator Card. 

The various points of the stroke at which important events occur,, 
important in their pressure-volume significance, have names, as do also the| 
lines between the points, and these names are more or less commonly accept 
and generally understood as follows: letters referring to the diagram Fig. 52. 

Point Names: Events of Cycle. 

A. Admission is that point of the stroke where the supply valve i$ 

opened. 

B. Cut-off is that point of the stroke where the supply valve is closed. 

C. Release is that point of the stroke where the exhaust valve is opened. 

D. Compression is that point of the stroke where the exhaust vab 

is closed. 
Names of Lines, or Periods: 

i-B. Admission or steam line joins the points of admission and cut-off] 

B-C. Expansion line joins the points of cut-off and release. 

C-D. Exhaust line joins the points of release and compression if th< 

is any, or admission if there is not. 
D-A. Compression line joins the points of compression and admission 



WORK OF PISTON ENGINES 



191 



By reason of the interferences discussed, these points on actual indicator 
ards may be difficult to locate, one line merging into the next in so gradual 
manner as to make it impossible to tell where the characteristic point lies, 
^s will be apparent from Fig. 53, in which is reproduced a number of actual 
ndicator cards. In such cases equivalent points must be located for study 








KG. 53. Actual Steam Engine Indicator Cards Showing Distortion of Lines and Uncertain 

Location of Characteristic Points. 






These same terms, which it appears sometimes refer to points and some- 
Imes to lines, are also used in other senses, for example, cut-off is com- 
Qonly used to mean the fraction of stroke completed up to the point of 
ut-off, and compression that fraction of stroke remaining incomplete at the 
int of compression, while compression is also sometimes used to express 
.he pressure attained at the end of the compression line. In general, there 



192 



ENGINEERING THERMODYNAMICS 



is nothing in the use of these words to indicate just which of the various 
meanings is intended except the text, and experience will soon eliminate most 
of the possible chances of confusion. 

Prob. 1. Draw a card in which admission and exhaust are late. Draw a card in 
which there is no compression, in which compression is very early, so that compression 
pressure is equal to admission pressure. Draw a card with per cent cut-off, and cut 
off = 100 per cent. Draw cards with same cut-off but with varying initial pressures. 

Prob. 2. The following diagrams, Fig. 54, are reproductions of indicator care 
actually taken from engines. Explain what features are peculiar to each and 
possible give an explanation of the cause. 






No. 3 



No. 4 



No. 5 





No. 7 





No. 8 



Fi<;. 54. Indicator Diagram from Steam Engines with Improperly Set Valve Gear. 

For example, in No. 1 a line of pressure equalization between the end of the com 

sion line and the beginning of the admission line inclines to the right instead ol 

being perpendicular, as in a perfect diagram. This is due to the fact that admis- 

ion does not occur until after the piston has begun to move outward, so that pres- 

t occur at constant volume, but during a period of increasing volume 

2. Standard Reference Cycles or PV Diagrams for the Work of Expansive 

Fluids in a Single Cylinder. Simple Engines. To permit of the derivatioi 

: a formula for the work of steam, compressed air, or any other fluid in 



WORK OF PISTON ENGINES 193 

jylinder, the various pressure volume, changes must be defined algebraically. 
The first step is, therefore, the determination of the cycle or pressure-volume 
liagram representative of the whole series of processes and consisting of a 
lumber of well-known phases or single processes. These phases, ignoring all 
sorts of interferences due to leakage or improper valve action, will consist of 
constant-pressure and constant-volume lines representing fluid movement 
into or from the cylinder, combined with expansion and compression lines 
representing changes of condition of the fluid isolated in the cylinder. These 
expansion and compression lines represent strictly thermal phases, laws for 
which will be assumed here, but derived rigidly later in the part treating of 
the thermal analysis; however, all cases can be represented by the general 
expression 

PV* = C, 

in which the character of the case is fixed by fixing the value of s. For all 
gases and for vapors that do not contain liquid or do not form or evaporate 
any during expansion or compression, i.e., continually superheated, the exponent 
: s may have one of two characteristic values. The first is isothermal expansion 
and compression, and for this process s is the same for all substances and 
equal to unity. The second is for exponential expansion or compression and 
for this process s will have values peculiar to the gas or superheated vapor 
under discussion, but it is possible that more than one substance may have the 
Same value, as may be noted by reference to Section 8, Chapter I, from 
which the value s = 1.406 for air and s = 1.3 for superheated steam or ammonia 
adiabatically expanding are selected for illustration. , 

When steam or any other vapor not so highly superheated as to remain 
tree from moisture during treatment is expanded or compressed in cylinders 
different values of s must be used to truly represent the process and, of course, 
there can be no isothermal value, since there can be no change of pressure of 
wet vapors without a change of temperature. For steam expanding adiabatically 
f he value of s is not a constant, as will be proved later by thermal analysis, so 
that the exact solution of problems of adiabatic expansion of steam under 
ordinary conditions becomes impossible by pressure-volume analysis and can 
be handled only by thermal analysis. However, it is sometimes convenient 
or desirable to find a solution that is approximately correct, and for this a 
sort of average value for s may be taken. Rankine's average value is s = 1 . 1 1 1 = l f- 
jfor adiabatic expansion of ordinarily wet steam, and while other values have 
ibeen suggested from time to time this is as close as any and more handy than 
most. The value s = 1.035+.14X(the original dryness fraction), is given by 
Perry to take account of the variation in original moisture. 

Steam during expansion adiabatically, tends to make itself wet, the 
Condensation being due to the lesser heat content by reason of the work done; 
but if during expansion heat be added to steam originally just dry, to keep it so 
continuously, as the expansion proceeds, it may be said to follow the saturation 
law of steam, for which s = 1.0646. This is a strictly experimental value found 



194 



ENGINEERING THERMODYNAMICS 











r 






































s 


s^_ 


Exp. 






\r 


=1 

Ex 


,. 






A 












B 


^s 


"I 




<\ 


























I 






^ 


^, 

































Exp. 



Exp 



Exp. 




1 








r 












| 


I 










-. 


t' 




X 


r-S= 
S^E 


tp. 




-8 = 


1 










I 


8 = 


J 


5 


^i 




s --Comp. 












^-c< 


mp. 




1 

















[ 








r 

CI 

j=l 












P 


=1 

-Exp. 






1 


-s-l 








I 


s 








! 


\- 


Exp. 






\ 


.*-Coi 


pp>- 


1 




V 

V 


^ 


- ^. 


~~ _" 











,c- 



c 



M 



Com] 



f 








f 












\ 


\rX- 







Exp. 






\ 


V 


N 


' 


"1 


\ 


\X 




L 




8 




x^- 


Com] 


) 


1 


Ck 


mp. 


Jl 





= - 











te 



Exp. 




FIG. 55. Standard Reference Cycles or Pressure-volume Diagrams for Expansive Fluid 

in Simple Engines. 



WORK OF PISTON ENGINES 



195 



by studying the volume occupied by a pound of just dry steam at various 
pressures, quite independent of engines. 

Direct observation of steam engine indicator cards has revealed the fact 
that while, in general, the pressure falls faster at the beginning of expansion 
and slower at the end than would be the case if s = l, yet the total work is 
about the same as if s had this value all along the curve. This law of expansion 
and compression, which may be conveniently designated as the logarithmic law, 
Js almost universally accepted as representing about what happens in actual 
fetaam engine cylinders. Later, the thermal analysis will show a variation of wet- 
ness corresponding to s = l, which is based on no thermal hypothesis what- 
ever, but is the result of years of experience with exact cards. Curiously 



p 

p 




a 




















a 






















V 


-EXE 


.PV= 


c 






p 








V 


-Exp. 


pys= 


c 






e 








\ 


b 
















s 


\ 










CYCLE 1 \. 

SIMPLE ENGINE LOGARITHMIC 
EXPANSION. 
ZERO CLEARANCE 


^ 


**Nfc^^ 






e CYCLE 2 ""^^ 

SIMPLE ENGINE EXPONENTIAL 
EXPANSION. 
ZERO CLEARANCE 












C 






"^ > 


C 
















































d 
















d 












a 














V 






a 














V 










V 


-Exp. 


PV= 








p 


I/ 






V 


Exp. 


?vs= 


3 








/ 


Com] 


).PV= 


\ 

c 


ss 










E 


-Com 


).PYS 


=c X 


V 








- 


I CYCLES ^\ 

A SIMPLE ENGINE LOGARITH 
\EXPANSION AND COMPRESS! 
^K WITH CLEARANCE 


viicT^ 

ON. 








V 


CYCLE 4 ^^ 

IMPLE ENGINE EXPONENT! 
EXPANSION AND COMPRESS 
V WITH CLEARANCE 


fr-^. 








c 




1 


ION. 


** 


c 




\^ 




















X^ 






















d 














d 









FIG. 56. Simple Engine Reference, Cycles or PV Diagrams. 

enough, this value of s is the same as results from the thermal analysis of con- 
stant temperature or isothermal expansion for gases, but it fails entirely to 
represent the case of isothermal expansion for steam. That s = 1 for isothermal 
gas expansion and actual steam cylinder expansion is a mere coincidence, a 
fact not understood by the authors of many books often considered standard, 
as in them it is spoken of as the isothermal curve for steam, which it most 
positively is not. This discussion of the expansion or compression laws indicates 
that analysis falls into two classes, first, that for which s = l, which yields a 
logarithmic expression for work, and second, that for which s is greater or 
less than one, which yields an exponential expression for work, and the former 
will be designated as the logarithmic and the latter as the exponential laws, for 
convenience. 



196 



ENGINEERING THERMODYNAMICS 



The phases to be considered then may be summed up as far as this analysis 
is concerned as: 

1. Admission or exhaust, pressure constant, P = const. 

2. Admission or exhaust, volume constant, F = const. 

3. Expansion, PV = const., when s = l. 

4. Expansion, PV* = const., when s is greater or less than 1. 

5. Compression, PV = const., when s = l.' 

6. Compression, PV S = const., when s is greater or less than 1. 

Considering all the possible variations of phases, there may result any or 
the cycles represented by Fig. 55. These cycles have the characteristics indicated 
by the following table, noting the variation in the law of the expansion or 
compression that may also be possible. 



Cycle. 


Clearance. 


Expansion. 


Compression. 


A 

B 
C 
D 


Zero ' 
Zero 
Zero 
Zero 


Zero 
Little 
Complete 
Over-expansion 


Zero 
Zero 
Zero 
Zero 


E 
F 
G 
H 


Little 
Little 
Little 
Little 


Zero 
Little 
Complete 
Over-expansion 


Zero 
Zero 
Zero 
Zero 


I 
J 
K 
L 


Little 
Little 
Little 
Little 


Zero 
Little 
Complete 
Over-expansion 


Little 
Little 
Little 
Little 


M 
N 
O 
P 


Little 
Little 
Little 
Little 


Zero 
Little 
Complete 
Over-expansion 


Complete 
Complete 
Complete 
Complete 


Q 

R 

S 
T 


Little 

Little 
Little 
Little 


Zero 
Little 
Complete 
Over-expansion 


Too much 
Too much 
Too much 
Too much 



It is not necessary, however, to derive algebraic expressions for all these 
cases, since a few general expressions may be found involving all the variables 
in which some of them may be given a zero value and the resulting expression 
will apply to those cycles in which that variable does not appear. The result- 
ing cycles, Fig. 56, that is is convenient to treat are as follows: 

SIMPLE ENGINE REFERENCE CYCLE OR PV DIAGRAMS 

CYCLE 1. Simple Engine, Logarithmic Expansion without Clearance. 
Phase (a) Constant pressure admission. 

(6) Expansion PV = const, (may be absent). 

(c) Constant volume equalization of pressure with exhaust (may be absent). 

(d) Constant pressure exhaust. 

(e) Constant (zero) volume admission equalization of pressure with supply. 



WORK OF PISTON ENGINES 



197 



YCLE II. Simple Engine, Exponential Expansion without Clearance. 
Phase (a) Constant pressure admission. 

(6) Expansion P V s = const, (may be absent). 

(c) Constant- volume equalization of pressure with exhasut (may be absent). 

(d) Constant-pressure exhaust. 

" (e) Constant (zero) volume admission equalization of pressure with supply. 

YCLE III. Simple Engine, Logarithmic Expansion and Compression with Clearance. 
Phase (a) Constant pressure admission. 

" (6) Expansion PV = const, (may be absent). 

(c) Constant volume equalization of pressure with exhaust (may be absent). 

(d) Constant pressure exhaust. 

(e) Compression PV = const, (may be absent). 

(/) Constant volume admission, equalization of pressure with supply (may 
be absent). 

YCLE IV. Simple Engine, Exponential Expansion and Compression with Clearance. 
Phase (a) Constant pressure admission. 

" (6) Expansion P V s = const, (may be absent). 

(c) Constant volume equalization of pressure with exhaust (may be absent). 
" (d) Constant pressure exhaust. 

(e) Compression PV S = const, (may be absent). 

" (/) Constant admission, equalization of pressure with supply (may be 
absent). 






P 

A 
57 






?n 








B 
















(in.pr) 

(rel.pr.) 
(bk.pr.) 
























\ 




























\ 


\ 




























> 


\ 






























\ 


^ 






























V ^ 


\ 




C 






























































E 




























D 












































-D 












1 
~^H 


G 












p 












Work of Expansive Fluid in Single Cylinder with No Clearance. 


V 

Logarithmic 



Expansion for Cycle I. Exponential for Cycle II. 

3. Work of Expansive Fluid in Single Cylinder without Clearance. Loga- 
ithmic Expansion, Cycle I. Mean Effective Pressure, Horse-power and 
onsumption of Simple Engine. Referring to the diagram, Fig. 57, the net 
wk, whether expansion be incomplete, perfect, or excessive, is the sum of 



198 



ENGINEERING THERMODYNAMICS 



admission and expansion work less the back-pressure work, or by are 

net work area, ABODE = admission work area ABFG 
+ expansion work area FBCH 
back-pressure work area GEDH 

or algebraically, 



- 



%r-P*V 



(230) 



Work of cycle in foot-pounds is 



-P,V, (a) 

-P*V d (b) 

. (231) 

(c) 

(<D 



As Vd or V e represent the whole displacement, the mean effective pressure 
will be obtained by dividing Eqs. (231), by V d or V e , giving, 



=P. 

m.e.p. = PC 



(232) 



Pt (<*) 
P (6) 

p d (d) 
-P^ (e) 



Similarly, dividing the Eq. (232) by the volume of fluid admitted, V b , will 
give the work per cubic foot, which is a good measure of economy, greatest 
economy being defined by maximum work per cubic foot, which, it may be noted, is 
the inverse of the compressor standard. 



Work per cu.ft. supplied = P/l+loge } -P d ~ (a) 

\ "/ V b 



=r')-P, (6) 



(233) 






WOEK OF PISTON ENGINES 



199 



According to Eq. (19), Chapter I, the piston displacement in cubic feet 

13 750 
>er hour per I.H.P. is for z = l>7 r]n J , and this multiplied by the fraction of 

yhole displacement occupied in charging the cylinder or representing admission, 
fhich is 

V* V, 

V d r TV 

ll give the cubic feet oj high pressure fluid supplied per hour per I.H.P., 
lence 



:u.ft.supplied per hr.per I.H.P. = 13 > 750 



TF () 



_ 13,750 P c 
~(m.e.p.) X P 6 (b) 



(234) 



Introducing a density factor, this can be transformed to weight of fluid. If 
phen #1 is the density of the fluid as supplied in pounds per cubic foot, 



Lbs. fluid supplied per hr. per I.H.P. = 



13,750 



13,750 



. . . (235) 



these expressions, Eqs. (230) to (235), for the work of the cycle, the 
nean effective pressure, work per cubic feet of fluid supplied, cubic feet and 
xmnds of fluid supplied per hour per I.H.P., are in terms of diagram point 
jonditions and must be transformed so as to read in terms of more generally 
lefined quantities for convenience in solving problems. The first step is to 
ntroduce quantities representing supply and back pressures and the amount 
)f expansion, accordingly: 

Let (in.pr.) represent the initial or supply pressure p b expressed in pounds 
per square inch; 

(rel.pr.) represent the release pressure p c , in pounds per square inch; 

(bk.pr.) represent the back pressure pa, in pounds per square inch; 
" R v represent the ratio of expansion defined as the ratio of largest to 



smallest volume on the expansion line ( ^J, or f ~] which is, of 
course, equal to the ratio of supply to release pressure ( ) , when 

\Pc / 

the logarithmic law is assumed; 

D represent the displacement in cubic feet which is V d or V c when no 
clearance is assumed; 



200 



ENGINEERING THERMODYNAMICS 



Let Z represent the fraction of stroke or displacement completed up 
cut-off so that ZD represents the volume Vt> admitted to the cylinde 

In this case when clearance is zero, Z=-^ , 



Work of the cycle in foot-pounds 



W=l44\(in.pT.) l+lo ^ Rv -(bk. P T.)]D (a) 1 
L #rj J 

= 144[(rel.pr.) (1 +loge Rv) (bk.pr.)JD (6) J 
m.e.p. = (rel.pr.) (1 +log e R v ) (bk.pr.) (a) 

(bk.pr.) (6) 

= (in.pr.)zf 1 +log e ) - (bk.pr.) (c) 



Work per cu.ft. supplied = 144[(in.pr.) (1 -f-loge R v ) - (bk.pr.)^ F ] (a) 

.(bk^rOl 

JJ 



1 q 7 c A i 

Cu.ft. supplied per hr. per I.H.P. = ^^ ~ (a) 



13,750 
(m.e.p.) 



Lbs. supplied per hr. per I.H.P. 



13,750 Si ,, 
(m.e.p.) R v ( 



13,750 
(m.e.p.) 



ZSi(b) 



(236 



(237 



(239 



(240 



The indicated horse-power may be found by multiplying the work of th 
cycle, Eq. (236), by the number of cycles performed per minute n and divid 
ing the product by 33,000. 



H^o, (241 



or 



THP = 



229.2 



(242 



In any of these expressions where R v is the ratio of greatest to smallest volume 

during expansion, either R P , ratio of greater to smaller pressures, or 1, the 

& 



WORK OF PISTON ENGINES 



201 



reciprocal of the cut-off, may be substituted, since the expressions apply only 
to the logarithmic law, and clearance is assumed equal to zero. When 
clearance is not zero; it is shown later that the cut-off as a fraction of stroke 
is not the reciprocal of R P or R v . 

These expressions are perfectly general, but convenience in calculation 
will be served by deriving expressions for certain special cases. The first of 
these is the case of no expansion at all, the second that of complete expansion 
without over-expansion. This latter gives the most economical operation from 
the hypothetical standpoint, because no work of expansion has been left 
unaccomplished and at the same time no negative work has been introduced 
by over-expansion. 



k. pr.) 



FIG. 58. First Special Case of Cycles I and II. Expansion = zero. Full Stroke Admission. 

First Special Case. If there is no expansion, together with the above assump- 
tion of no clearance, the diagram takes the form (Fig. 58), and 

TP=144D[(in.pr.)- (bk.pr.)] . .' 

m.e.p. = (in.pr.) (bk.pr.). . . . 

Work per cu.ft. supplied = 144[(in.pr.) (bk.pr.)]. . . 

Cu.ft. supplied per hr. per I.H.P. =-r- \' /ui \- 

(in.pr.) (bk.pr.) 

13,750Bi 
Lbs. supplied per hr. per I.H.P. = ^- -^-7^--^. - 



...... (243) 

. . . . . (244) 

. . . . . (245) 

(246) 



(247) 



202 



ENGINEERING THERMODYNAMICS 



Second Special Case. When the expansion is complete without over-expan- 
sion, no clearance, the points C and D, Fig. 59, coincide, and (rel.pr.) = (bk.pr.), 

hence R v =Rp= /V?' pr -4-==~- This value of cut-off, Z, is known as best cut-off, as 
(bk.pr.) 



U is that which uses all the available energy of the fluid by expansion. 



(m.e.p.) = (in.pr.) 



Work per cu.ft. supplied ==144(in.pr.) 



loge Ri 
R P 



(248) 

(249) 
(250) 




D (rel.pr.)= 
' (bk.pr) 



If . 

FIG. 59. Second Special Case of Cycles I and II. Complete Expansion Without Over 

Expansion Case of Best Cut-off. 



Cu.ft. supplied per hr. per I.H.P.= 13 ' 7 1 50 - . 

(in.pr.) loge R P 



Lbs. supplied per hr. per I.H.P. = ~ 

(in.pr.) loge R 



(251) 
(252) 



Example 1. Method of calculating diagrams. Fig. 57 and Fig. 59. 
Assumed data for Fig. 57. 

Pa=P u = 90 Ibs. per sq.in. abs. V a = V e = cu.ft. 
Pa =P = 14 Ibs. per sq.in. abs. V c = V d = 13.5 cu.ft. 

F ft = 6cu.ft. 






WORK OF PISTON ENGINES 203 

To obtain point C: 

y b g 

PC =Pb X =90 X =40 Ibs. per sq.in. abs. 

V c lo.O 

Assumed data for Fig. 59. 



P a =Pb= 90 Ibs. per sq.in. V a = V e = cu.f t. 
P d =Pe = 14 Ibs. per sq.in. V d = 13.5 cu.ft. 



To obtain point B: 



^ = 13.5 X =2.1 cu.ft. 

JT6 90 



Example 2. A simple double-acting engine admits steam at 100 per square inch 
absolute for \ stroke, allows it to expand to the end of the stroke and then exhausts it 
against a back pressure of 5 Ibs. per square inch absolute. If the engine has no 
learance, a 7 X9-in. cylinder and runs at 300 R.P.M., what is the horse-power and steam 
onsumption when steam is expanding according to the logarithmic law? Note: 
cu.ft. steam at 100 Ibs. per square inch absolute weighs .2258 Ib. 

From Eq. (237 6 ), 



-5=54.7: 



4 4 



_ (m.e.p.)Lan 54.7 X. 75x38.5x600 
33,000 33,000 



directly from Eq. (242) 

I H P = 



229.2 

.2X600X54.7 
229.2 

Lbs. steam per I.H.P. = - - X , 
m.e.p. R v 



=28, 



=1413 



Therefore, steam per hour used by engine = 14.15x28 =396 Ibs. 

Prob. 1. A steam engine has a cylinder 12x18 ins. with no clearance. It runs at 
200 R.P.M. and is double-acting. If the steam pressure be fixed at 100 Ibs. per 
square inch absolute, and the back pressure at 10 Ibs. per square inch abs., show how 
the horse-power and steam consumption will vary as cut-off increases. Take cut-off 
from i to f by eighths. Plot. 



204 ENGINEEEING THERMODYNAMICS 

Prob. 2. Two engines of the same size and design as above are running on a steam 
pressure of 100 Ibs. per square inch absolute, but one exhausts through a long pipe to 
the atmosphere, the total back pressure being 20 Ibs. per square inch absolute, 
while the other exhausts into a condenser in which the pressure is but 3 Ibs. per 
square inch absolute. If the cut-off is in each case f , how will the I.H.P. and steam 
used in the two cases vary? 

Prob. 3. By finding the water rate and the horse-power in the two following cases, 
show the saving in steam and loss in power due to using steam expansively. A pump 
having a cylinder 9 X 12 ins. admits steam full stroke, while an engine of same size admits 
it but i of the stroke; both run at the same speed and have the same back pressure. 

Prob. 4. Steam from a 12 X24 in. cylinder is exhausted at atmospheric pressure 
(15 Ibs. per square inch absolute) into a tank, from which a second engine takes steam. 
Neither engine has clearance. The first engine receives steam at 100 Ibs. per square 
inch absolute and the cut-off is such as to give complete expansion. The second engine 
exhausts into a 24 in. vacuum and its cut-off is such that complete expansion occurs in 
its cylinder. Also the cylinder volume up to cut-off equals that of the first cylinder 
at exhaust. If the stroke is the same in both engines and the speed of each is 200 
R.P.M., what is the diameter of the larger cylinder, the total horse-power developed, 
the total steam used, and the work per cubic foot of steam admitted to the first 
cylinder, the water rate of each engine and the total horse-power derived from each 
pound of steam? 

Prob. 5. The steam pressure for a given engine is changed from 80 Ibs. per square 
inch gage to 120 Ibs. per square inch gage. If the engine is 12x16 ins., running 250 
R.P.M. with a fixed cut-off of 25 per cent and no clearance, the back pressure being 
15 Ibs. per square inch absolute, what will be the horse-power and the water rate in 
each case? 

NOTE: 1 cu.ft. of steam at 80 and 120 Ibs. weighs .215 and .3 Ib. respectively. 

Prob. 6. By trial, find how much the cut-off should have been shortened to 
keep the H.P. constant when the pressure was increased and what effect this would 
have had on the water rate. 

Prob. 7. A certain type of automobile engine uses steam at 600 Ibs. per square 
inch absolute pressure. The exhaust is to atmosphere. For a cut-off of | and no 
clearance, what would be the water rate? 

NOTE: for 600 Ibs. <?i=1.32. 

Prob. 8. Engines are governed by throttling the initial pressure or shortening the 
cut-off. The following cases show the effect of light load on economy. Both engines, 
12x18 ins., running at 200 R.P.M., with 125 Ibs. per square inch absolute. Initial 
pressure and back pressure of 10 Ibs. per square inch absolute. The load is sufficient 
to require full steam pressure at \ cut-off for each engine. Load drops to a point 
where the throttle engine requires but 50 Ibs. per square inch absolute initial pressure 
with the cut-off still fixed at . What is the original load and water rate, and new 
load and water-rate for each engine? 

NOTE: a for 125 Ibs. absolute = .279 and for 50 Ibs. =.117 Ib. 

Prob. 9. The guarantee for a simple engine 18x24 ins., running at 200 R.P.M., 
states that the I.W.R. when cut-off is } will not exceed 15 Ibs. if the initial pressure be 
100 Ibs. per square inch gage, and back pressure 5 Ibs. per square inch absolute. 
If engine has no clearance, see if this would be possible. 

4. Work of Expansive Fluid in Single Cylinder without Clearance. Expo- 
nential Expansion Cycle II. Mean Effective Pressure, Horse-power and 



WORK OF PISTON ENGINES 



205 



Consumption of Simple Engines. Referring to the diagram, Fig. 57, the 
work is given by the same areas as for Cycle I, Hut its algebraic expression 
is different because s is greater than 1 and an exponential expansion results on 
integration instead of a logarithmic one. 

In general, from Eq. (13a), Section 7, Chapter I, 



v 



Putting this in terms of initial conditions by the relations 

/PA /V c \* * 

\^p) = (~v) = #F and F < =F <* = 

\* / \ ' b/ 



Also 



there results 



W = P*V*+, _ P *^ b 1 (Rv s - l -l)-P d V^ 



, 

{S 



- 1 -i) - 



= 144D \Z (in.pr.) ^^p - (bk.pr.)l 



(c) 



which is the general equation for work of this cycle. 
Dividing by 7 6 , the volume of fluid supplied, 



Work per cu.ft. supplied = . 



_ 1 -P d Rv (a) 









(253) 



(254) 



(255) 



Similarly, the mean effective pressure results from dividing the work by the 
displacement, V d =Vi,Rv 



or 



(a) 



-J-tbk.pr.) (6) 



. (256) 



(c) 



206 



ENGINEERING THERMODYNAMICS 



First special case of no expansion, when /iV = l> results the same diagran 
as in the previous section, Fig, 58, and exactly the same set of formulas. 

Second special case when the expansion is complete without over-expansion 
is again represented by Fig. 59 and for it 



Whence 



or 



Work for complete expansion is 




R 



(6) 
() 



. . (257 



(D \ 
which is the general equation for the work of V b or ( ^r ) cubic feet of fluic 

when the economy is best or for best cut-off. 

The work per cubic foot of fluid supplied for this case of complete expansioi 
gives the maximum value for Eq. (255) and is obtained by dividing Eq. (257 



Max. work per cu.ft. supplied 



, -.(258 



which is the general equation for maximum work per cubic foot of fluid supplied 
The expression for mean effective pressure becomes for this case of best cut-off, 



or, 



. . . (259 



It is convenient to note that in using Eqs. (257), (258) and (259) it may b 
desirable to evaluate them without first finding R v . Since 



Vc 

v 



P. 



=p 



WORK OF PISTON ENGINES 207 

}his substitution may be made, and 

s-l 

Z? 7"> 

fly lp * 

(Example. Compare the horse-power and the steam consumption of a 9x12 in. 
simple double-acting engine with no clearance and running at 250 R.P.M. when initial 
pressure is 100 Ibs. per square inch absolute and cut-off is J, if 
(a) steam remains dry and saturated throughout expansion, 
(6) remains superheated throughout expansion, and 
(c) if originally dry and suffers adiabatic expansion. 
Back pressure is 10 Ibs. per square inch absolute. 



For case (a) . -1.0646 and (m.e.p.) -~ .w -10 =48.6; 

" (6) S = 1.3 and (m.e.p.) J?(!|_L-) -10=43.5; 
(c). -1.111 and fr*-PJ-T(lir-.iilx4-m) -10=47.5. 



/. I.H.P. for case (a) =46.9, 
(6) =42.0, 
(c) =45.8. 

From Eq. (240), Ibs. steam per hour per I.H.P. =.X- 

m.e.p. 

"I Q *7 PCn 5s 

.*. For case (a) steam per hr. =46.9 X ' . X-T 5 
i 48.o 4 



" (6) steam per hr. =42 X X j; 



" (c) steam per hr. =45.8 Xi X. 

47.5 4 

Prob. 1. On starting a locomotive steam is admitted full stroke, while in running 
he valve gear is arranged for f cut-off. If the engine were 18x30 ins., initial pressure 
150 Ibs. per square inch absolute, back pressure 15 Ibs. per square inch absolute, what 
would be the difference in horse-power with the gear in normal running position and 
in the starting position for a speed of 20 miles per hour with 6-ft. driving wheels? Con- 
sider the steam to be originally dry and expanding adiabatically. What would be the 
difference in steam per horse-power hour for the two cases and the difference in total 
steam? Clearance neglected. 

Prob. 2. Consider a boiler horse-power to be 30 Ibs. of steam per hour; what must be 
the horse-power of a boiler to supply the following engine? Steam is supplied in a super- 



208 ENGINEERING THERMODYNAMICS 

heated state and remains so throughout expansion. Initial density of steam = .21 Ibs. per 
cubic foot. Engine is 12x20 ins., double-acting, 200 R.P.M., no clearance, initial 
pressure 125 Ibs. per square inch absolute, back pressure a vacuum of 26 ins. of 
mercury. Cut-off at maximum load f, no load, f&. What per cent of rating of boiler 
will be required by the engine at no load? 

Prob. 3. While an engine driving a generator is running, a short circuit occurs 
putting full load on engine, requiring a f cut-off. A moment later the circuit-breaker 
opens and only the friction load remains, requiring a cut-off of only . The engine 
being two-cylinder, double-acting, simple, 12 X18 ins., running at 300 R.P.M., and having 
no clearance, what will be the rate at which it uses steam just before and just after 
circuit-breaker opens if the steam supplied is at 125 Ibs. per square inch absolute and is 
just dry, becoming wet on expanding, and back pressure is 3 Ibs. per square inch 
absolute? 

Prob. 4. A pumping engine has two double-acting steam cylinders each 9X12 ins. 
and a fixed cut-off of \. It runs at 60 R.P.M. on 80 Ibs. per square inch absolute steam 
pressure and atmospheric exhaust. Cylinder is jacketed so that steam stays dry 
throughout its expansion. How much steam will it use per hour? Neglect clearance. 

Prob. 6. If an engine 10x14 ins. and running 250 R.P.M. has such a cut-off that 
complete expansion occurs for 90 Ibs. per square inch absolute initial pressure, and at 
atmospheric (15 Ibs. absolute) exhaust, what will be the horse-power and steam used per 
hour, steam being superheated at all times, and what would be the value for the horse- 
power and steam used if full stroke admission occurred? 

Prob. 6. The steam consumption of an engine working under constant load is 
better than that of a similar one working under variable load. For a 16 X 24 ins. engine 
running at 250 R.P.M. on wet steam of 125 Ibs. per square inch absolute and atmospheric 
exhaust, find the horse-power and steam used per horse-power per hour for best con- 
dition and by taking two lighter and three heavier loads, show by a curve how steam 
used per horse-power per hour will vary. 

Prob. 7. For driving a shop a two cylinder single-acting engine, 6x6 ins., running 
at 430 R.P.M., is used. The cut-off is fixed at \ and intitial pressure varied to control 
speed. Plot a curve between horse-power and weight of steam per hour per horse- 
power for 20, 40, 60, 80, 100, 120 Ibs. per square inch absolute initial pressure and 
atmospheric exhaust. Steam constantly dry. Clearance zero. 

NOTE: <*i for above pressures equals .05, .095, .139, .183, .226, and .268 Ibs. per 
cubic foot, respectively. 

Prob. 8. Taking the loads found in Prob. 7, find what cut-off would be required j 
to cause the engine to run at rated speed for each load if the initial steam pressure 
were 100 Ibs. per square inch absolute, and the back pressure atmosphere, and a plot 
curve between horse-power and steam used per horse-power hour for this case. 

Prob. 9. For working a mine hoist a two-cylinder, double-acting engine is used 
in which compressed air is admitted f stroke at 125 Ibs. per square inch absolute and then 
allowed to expand adiabatically and exhaust to atmosphere. If the cylinders are 18 X24 
ins. and speed is 150 R.P.M., find the horse-power and cubic feet of high pressure air 
needed per minute. 

5. Work of Expansive Fluid in Single Cylinder with Clearance. Loga- 
rithmic Expansion and Compression; Cycle III. Mean Effective Pressure, 
Horse-power, and Consumption of Simple Engines. As in previous cycles, 
the net work of the cycle is equal to the algebraic sum of the positive work 






WORK OF PISTON ENGINES 



209 



one on the forward stroke and the negative work on the return stroke. By 
Teas, Fig. 60, this is 



Work area = JABN+NBCW- WDEO - OEFJ. 
Expressed in terms of diagram points this becomes 



W 



(260) 



(CD) 


c 


upply- 


Yolun 
7-R - 


e 


,- 


B 




in.pr. 












"V 


ft- 














Q 














\ 
















i 
















\ 














i 
















X 


\ 














F 


















\ 


^ 
































"^ 


^ 


-^C(t 




\ 






























\ 


\ 


























.__ 


.. 


\ 
























D( 


Ib 
























1 






1 
























"M 

|W 


j 


!c 










N 


D 












K-X-D^I V 



l.pr) 



k.pr) 



. 60. Work of Expansive Fluid in Single Cylinder with Clearance. Logarithmic Expan- 
sion for Cycle III. Exponential for Cycle IV. 

Expressing this in terms of displacement, in cubic feet D; clearance as a frac- 
iion of displacement, c; cut-off as a fraction of displacement, Z; compres- 
sion as a fraction of displacement, X; initial pressure, in pounds per square 
nch (in.pr.), and exhaust or back pressure, in pounds per square inch (bk.pr): 



P 6 = 144(in.pr.); 
P d = 144 (bk.pr.); 
(Vt > -V a )=ZD' ) 



D(Z+c) Z+c' 



V e = D(X+c). 

V e = D(X+c) = X+c 
V f DC c ' 



210 

Whence 
Work in ft.-lbs. per cycle is 

144D j (in.pr.) 

W 

-(bk.pr.) 1(1- 



ENGINEERING THERMODYNAMICS 



) l oge 



(261) 



From Eq. (261), the mean effective pressure, pounds per square inch, follows 
by dividing by 144D: 



(in.pr.) 
- (bk.pr.) (l- 



. . . (mean forward press.) 



) loge ^~t- c .J ( 



(262) 



mean bk.pr.) 



This is a general expression of very broad use in computing probable mean 
effective pressure for the steam engine with clearance and compression, or for 
other cases where it is practicable to assume the logarithmic law to hold. Fig. 
117, at the end of this chapter, will be found of assistance in evaluating this 
expression. 

Indicated horse-power, according to expressions already given, may be 
found by either of the following equations: 



33,000 



144(m.e.p.)Dn = (m.e.p.)Dn 
33,000 229.2~~' 






where L is stroke in feet, a is effective area of piston, square inches, n is the 
number of cycles performed per minute and D the displacement, cubic feet. 

It might seem that the work per cubic foot of fluid supplied could be 
found by dividing Eq. (261) by the admission volume, 



but this would be true only when no steam is needed to build up the pressure 
from F to A. This is the case only when the clearance is zero or when com- 
pression begins soon enough to carry the point F up to point A, i.e., when by 
compression the pressure of the clearance fluid is raised to the initial pressure. 
It is evident that the fluid supplied may perform the two duties: first, 
building up the clearance pressure at constant or nearly constant volume, 
and second, filling the cylinder up to cut-off at constant pressure. To measure 
the steam supplied in terms of diagram quantities requires the fixing of the 
volume of live steam necessary to build up the pressure from F to A and adding 
it to the apparent admission volume (V b V a ). This can be done by producing 



WOEK OF PISTON ENGINES 211 

,he compression line EF to the initial pressure Q, then LQ is the volume that 
ihe clearance steam would have at the initial pressure and QA the volume of 
ive steam necessary to build up the pressure. The whole volume of steam 
tdmitted then is represented by QB instead of AB or by (V b V g ) instead of 
^y (Vb Va), and calling this the supply volume, 



But 



Hence 

(Sup.Vol.) =D\(Z-\-c) (X-\-c)~- J-y , .... (263) 

which is the cubic feet of fluid admitted at the initial pressure for the dis- 
placement of D cubic feet by the piston. Dividing by D there results 



(264) 



which is the ratio of admission volume to displacement or cubic feet of live 
steam admitted per cubic foot of displacement. 

Dividing the work done by the cubic feet of steam supplied gives the 
economy of the simple engine in terms of volumes, or 

W 

Work per cu.ft. of fluid supplied = 



(Sup.Vol.) 
(in.pr.) \Z+ (Z+c)log e J^~| - (bk.pr.) ["(1 -X) + (X+c)lo& 



' (265) 



It is more common to express economy of the engine in terms of the weight 
of steam used per hour per horse-power or the " water rate," which in more 
general terms may be called the consumption per hour per I.H.P. 

Let di be the density or weight per cubic foot of fluid supplied, then the 
weight per cycle is (Sup.Vol.) di, and this weight is capable of performing 

W 

W foot-pounds of work or (Sup.Vol.) di Ibs.. per minute will permit of 

horse-power. But (Sup.Vol.) 3i Ibs. per mintue corresponds to 60 (Sup.Vol.) 
L Ibs. per hour, whence the number of pounds per hour per horse-power is 

60 (Sup.Vol.)fli 
TF/33,000 ' 



212 ENGINEERING THERMODYNAMICS 

which is the pounds consumption per hour per I.H.P., whence 

60X33,000(Sup.Vol.)3i 
Consumption in Ibs. per hr. per I.H.P.= ~~W~ *''' ' * ' ' 

which is the general expression for consumption in terms of the cubic feet of 
fluid admitted per cycle, ^i initial density, and the work per cycle. 

As work is the product of mean effective pressure in pounds per square foot, 
(M.E.P.,) and the displacement in cu.ft. or W = (M.E.P.)D, or in terms of 
mean effective pressure pounds per square inch TF = 144 (m.e.p.)D, the 
consumption may also be written 

60X33,000(Sup.Vol.)3i 
Consumption in lbs.,per hr. per I.H.P. = i44( m . e .p.)D 

13,750 (Sup.Vol.)3i 
(m.e.p.) D 

13,750 , , y . v(bk.pr.)1 

- * lt ..... (267) 



which gives the water rate in terms of the mean effective pressure, cut-off, 
clearance, compression, initial and back pressures and initial steam density. 
It is sometimes more convenient to introduce the density of fluid at the back 
pressure ^2, which can be done by the relation (referring to the diagram), 

p v _ PF or (in.pr.) _V e _di 
M^ ^ K ' 0r (bk.pr:) ^ S 

whence 



(bk.pr.) 



This on substitution gives 
Consumption in Ibs., per hr. per I.H.P. 



Since the step taken above of introducing ^2 has removed all pressure or 
volume ratios from the expression, Eq. (268) is general, and not dependent 
upon the logarithmic law. It gives the consumption in terms of mean effective 
pressure, cut-off, clearance, compression and the density of steam at initial 
and back pressure, which is of very common use. 

It cannot be too strongly kept in mind that all the preceding is true only when 
no steam forms from moisture water during expansion or compression or no steam 
condenses, which assumption is known to be untrue. These formulae are, there- 
fore, to be considered as merely convenient approximations, although they 



WORK OF PISTON ENGINES 



213 



ire almost universally used in daily practice. (See the end of this chapter for 
liagrams by which the solution of this expression is facilitated.; 

Special Cases. First, no expansion and no compression would result in 
Fig. 61. For it 

TF=144Z)[(in.pr.)- (bk.pr.)] (269) 

(m.e.p.) = (in.pr.)-(bk.pr.) . . (270) 



p 

L 


n 












Su] 


ply-V 


ft |. 1T _ 












_, 


B 


(in.pr.) 
(bk.pr.) 


j 






















JQ 






































































































































































cD 




































i 

i 
































N 


\ 






























D 


i 




























K 


p 

































w v 



FIG. 61. First Special Case of Cycles III and IV. Expansion and Compression both Zero, 

but Clearance Finite. 



The volume of fluid supplied per cycle is QB, or from Eq. (263) it is 



(271) 



13,750 f (bk.pr.)l, 

Consumption in Ibs. per hr. per I.H.P.= ^ -- N ,, , -- , 1 +cc- r ~ \oi 

(m.pr.) - (bk.pr.) |_ (in.pr.) ] 2 

or in terms of initial and final densities, 

13 750 

' 



(272) 



Consumption in Ibs. per hr. per l.H.P. = T . 



(273) 



The second special case is that of complete expansion and compression, as 
indicated in Fig. 62. Complete expansion provides that the pressure at the 



214 



ENGINEERING THERMODYNAMICS 



end of expansion be equal to the back pressure, and complete compression that 
the final compression pressure be equal to the initial pressure. 

Here 



and hence 



and 



_ = _ = X+c = (in.pr.) 

F 6 Fa Z+c ' c (bkpr.)' 



F c -F_in.pr. y \-p\i I c c (' m ' w '\\ 

V^Fa'bk^F.' ^L \bk.prJJ 



(V b -Va)=ZD. 






in.pr.) 




a~ (bk.pr.) 



FIG. 62. Second Special Case of Cycles III and IV. Perfect Expansion and Perfect 
Compression with Clearance. 

Hence by substitution 



ZD 



(bk.pr.) ' 



from which 



Again, 



v.-v t 



D 



-[(&H 



- 






WORK OF PISTON ENGINES 215 



5q. (274) gives the cut-off as a fraction of the displacement necessary to give 
complete expansion, while Eq. (275) gives the compression as a fraction of dis- 
jlacement to give complete compression, both in terms of clearance, initial 
Dressure and back pressure, provided the logarithmic law applies to expansion 
and compression. 

Substitution of the values given above in Eq. (261) gives, after simplifi- 
3ation, 

TF = 144D[(l+c)(bk.pr.)-c(in.pr.)]log e ^^ . . (276) 

...-.. (277) 



sin this case the volume supplied is exactly equal to that represented by the 
admission line AB, and is equal to 

(Sup.Vol.)=ZD. . . . (278) 

Hence, the consumption, in pounds fluid per hour per I.H.P. in terms of 
initial density, is 

13 750 

Consumption in Ibs. per hr. per I.H.P. = '- r Zd\ t 

(m.e.p.) 

but 

Z 



m.e.p. ,. . f_ , .(bk.pr.) 1, (in.pr.) ' X1 (in.pr.) 

(in.pr.) (\-\-c)-. c log e /U1 N (in.pr.) logeTrr^ : 

' [ (m.pr.) J 3 (bk.pr.) & e (bu^> 

hence 



13 ' tyv^ i 

Consumption in Ibs. fluid per hr. per I.H.P. = ^^ -r-. . . (279) 

/ \ i (m.pr.) 
(m.pr.) ^ v 



This last equation is interesting in that it shows the consumption (or water 
rate, if it is a steam engine) is independent of clearance, and dependent only 
upon initial density, and on the initial and final pressures. 

An expression may also be easily derived for the consumption in terms of 
initial and final density, but due to its limited use, will not be introduced here. 

Example 1. Method of calculating Diagrams Fig. 60 and Fig. 62. 
Assumed data for Fig. 60: 

p a =p a =p b = 90 Ibs. per square inch abs. V a = V f = .5 cu.ft. 
P g =p e =P d = 14 Ibs. per square inch abs. V d = V c = 135 cu.ft. 
P f = 50 Ibs. per square inch abs. F& = 6 cu.ft. 



216 ENGINEERING THERMODYNAMICS 

To obtain point C: 

p t =p b X =90X =40 Ibs. per sq.in. 

To obtain point E: 

7 e = F/X- / = .5x- = 1.78cu.ft. 

Pe 14 






To obtain point Q: 

P f 50 

Intermediate points from B to C and E to Q are found by assuming volumes 
and computing the corresponding pressures by relation P x V x =PbVb or P x V x =P e V e . 

Clearance is - -A=r =~ =3.8 per cent, 

Vd V a 16 

Cut-off is ft ~ a =-^ =42.3 per cent, 

F V a 1 28 
Compression is =- =-^r- =9.9 per cent. 

V d~ Va lo 

Assumed data for Fig. 62. 

p a =p 6 =90 Ibs. per square inch absolute. V a = .5 cu.ft. , 

P e = Pc = 14 Ibs. per square inch absolute. Vd = 13.5 cu.ft. 

To obtain point B: 

V b = V C ^= 13.5 X^: =2.11 cu.ft. 
To obtain point E: 

rr TT . .*a *J1J 



Intermediate points from B to C and from A to E are to be found by assuming 
various volumes and finding the corresponding pressures from relation P x V x =PaVa or 

p t v x =p b v b . 

Example. 2. What will be the horse-power of, and steam used per hour by the 
following engine: 

(a) cut-off 50 per cent, compression 30 per cent, 

(6) complete expansion and compression, 

(c) no expansion or compression. 

Cylinder, 12 Xl8-in. double-acting, 200 R.P.M., 7 per cent clearance, initial pressure 
85 Ibs. per square inch gage, back pressure, 15 Ibs. per square inch absolute, and 
logarithmic expansion and compression. 



WORK OF PISTON ENGINES 217 

NOTE: d for 85 Ibs. gage = .23, <?i for 15 Ibs. absolute = .038 cu.ft. 
(a) From Eq. (262) 



-(bk.pr.) (1 - 

log e -- 



=86 -20 =66 Ibs. sq.in. 

^ (m.e.p.) Lan ^66X1.5X113.1X400 
33,000 33,000 



From Eq. (267) steam per hour per I.H.P. in pounds is 
13,750 [., . /bk.prAl 

, -- r\(Z+c)-(X+c)\T -8l, 
(m.e.p.) |_ V 'Vm.pr./J ' 



Hence total steam per hour =25x1 35 =3380 Ibs. 
(6) From Eq. (277) 

(m.e.p.) =[(1 +c)(bk. pr.) -c(in. pr.)] lo ge , 



=[(1+.07) X15-.07X100] log e 6.67 = 17.2 Ibs. sq.in. 

17.2X1.5X113.1X400 
LH * R= ~^3,00^~ 

From Eq. (279) 

13,750 Si 13,750 X. 23 

Steam per I.H.P. per hour = - -/-V= Too~xl^ 



Total steam per hour = 16.6x35. 4 =588 Ibs. 

(c) From Eq. (270) (m.e.p.) =(in.pr.) -(bk.pr.) =100-15=85 Ibs. sq.in. 



Eq. (273) 

1 3 7^0 1 ^ 7^0 

Steam per I.H.P. per hour = ,^[(1 +c)*i -o 2 ] - ^-[1.07x.23 -.07X.038] =35.4. 

(m.e.p.) 85 

Total steam per hour = 174.5x35.4 =6100 Ibs. 

Prob. 1. What will be the horse-power and water rate of a 9x12 in. simple engine 
having 5 per cent clearance, running at 250 R.P.M. on 100 Ibs. per square inch abso- 



218 ENGINEERING THERMODYNAMICS 

lute initial pressure and 5 Ibs. per square inch absolute back pressure when the cut-off is 
, -J-, and |, expansion follows the. logarithmic law, and there is no compression? 

NOTE: 8 for 100 Ibs. absolute = .23, 8 for 5 Ibs. absolute = .014. 

Prob. 2. Will a pump with a cylinder 10x15 ins. and 10 per cent clearance give the 
same horse-power and have the same water rate as a pump with cylinders of the same 
size but with 20 per cent clearance, both taking steam full stroke? Solve for a case 
of 125 Ibs. per square inch absolute initial pressure, atmospheric exhaust and a speed of 
50 double strokes. No compression. 

NOTE: 8 for 125 Ibs. absolute = .283, 8 for 15 Ibs. absolute = .039. 

Prob. 3. Solve the above problem for an engine of the same size, using steam expan- 
sively when the cut-off is \ and R.P.M. 200, steam and exhaust pressure as in Prob. 2 
and compression zero. 

Prob. 4. Two engines, each 9x15 ins., are running on an initial pressure of 90 Ibs. 
per squareHnch absolute, and a back pressure of one atmosphere. One has no clearance, 
the other 8 per cent. Each is cutting off so that complete expansion occurs. The speed 
of each is 200 and neither has any compression. What will be the horse-power and 
water rate? 

NOTE: 8 for 90 Ibs. =.24, 8 for 15 Ibs. = .039. 

Prob. 5. By finding the horse-power and water rate of a 12x18 in. double-acting 
engine with 8 per cent clearance, running at 150 R.P.M. on an initial steam pressure 
of 90 Ibs. per square inch absolute and atmosphere exhaust for a fixed cut-off of \ and 
variable compression from to the point where the pressure at the end of compression 
is equal to 125 per cent of the initial pressure, plot the curves between compression and 
horse-power, and compression and water rate to show the effect of compression on 
the other two. 

NOTE: 8 for 90 Ibs. =.21, 8 for 15 Ibs. =.039. 

Prob. 6. A steam engine is running at such a load that the cu^-off has to be f at 
a speed of 150 R.P.M. The engine is 14 X20 ins. and has no clearance. Initial pressure 
100 Ibs. per square inch absolute and back pressure 5 Ibs. per square inch absolute. 
What would be the cut-off of an engine of the same dimensions but with 10 per cent 
clearance under similar conditions? 

Prob. 7. The steam pressure is 100 Ibs. per square inch gage and the back 
pressure is 26 ins. of mercury vacuum when the barometer is 30 ins. For a 16x22 
in.engine with 6 per cent clearance running at 125 R.P.M., cut-off at \ and 30 per cent 
compression, what will be the horse-power and the water rate? Should the steam 
pressure be doubled what would be the horse-power and the water rate? If it should 
be halved? 

NOTE: 8 for 100 Ibs. gage = .2017, 8 for 26 ins. Hg.=.0058. 

Prob. 8. While an 18x24 in. simple engine with 4 per cent clearance at speed 
of 150 R.P.M. is running with a \ cut-off and a compression of \ on a steam pressure of 
125 Ibs. per square inch gage, and a vacuum of 28 ins. of mercury, the condenser fails 
and the back pressure rises to 17 Ibs. per square inch absolute. What will be the change in 
the horse-power and water rate if all other factors stay constant? What would the new 
cut-off have to be to keep the engine running at the same horse-power and what 
would be the water rate with this cut-off? 

NOTE: 8 for 125 Ibs. gage =.315, 8 for 28 in. Hg.=.0029, 8 for 17 Ibs. absolute 
= .0435. 

Prob. 9. Under normal load an engine has a cut-off of |, while under light load 
the cut-off is but A. What per cent of the steam used at normal load will be used 



WOEK OF PISTON ENGINES 219 

it light load for the following data? Cylinder 10x18 ins.; 7 per cent clearance; 200 
R.P.M.; initial pressure 120 Ibs. per square inch gage; back pressure 2 Ibs. per square 
nch absolute; compression at normal load 5 per cent; at light load 25 per cent. 

NOTE: S for 120 Ibs. gage =.304, 8 for 2 Ibs. absolute = .0058. 
II 

6. Work of Expansive Fluid in Single Cylinder with Clearance ; Exponential 

Expansion and Compression, Cycle IV. Mean Effective Pressure, Horse- 

)ower and Consumption of Simple Engines. As pointed out in several places, 

;he logarithmic expansion of steam only approximates the truth in real engines 

rnd is the result of no particular logical or physically definable hypothesis as to 

}he condition of the fluid, moreover its equations are of little or no value 

r or compressed air or other gases used in engine cylinders. All expansions 

jjvhat can be defined by conditions of physical state or condition of heat, 

Including the adiabatic, are expressible approximately or exactly by a definite 

/alue of s, not unity, in the expression PV S = const. All these cases can 

hhen be treated in a group and expressions for work and mean effective 

|3ressure found for a general value of s, for which particular values belonging 

|x), or following from any physical hypothesis can be substituted. The area 

mder such expansion curves is given by Eq. (13) Chapter I, which applied to 

I:he work diagram, Fig. 60, in the same manner as was done for logarithmic 

:j 3xpansion, gives the net work: 

s-l 

(area JABCWJ) _ 

. (280) 



(area WDEFJW ) 

~~ L L \ v // J 
Introducing the symbols, 

P 6 = 144(in.pr.), V b = D(Z+c). 

P d = 144(bk.pr.), V e = D(X+c). 



(V-V a )=ZD, (T.rm 

(Vd -VJ=D(l-X), =-?' 



. . (281) 



Eq. (281) gives the work in foot-pounds for D cubic feet of displacement in a 
jylinder having any clearance c, cut-off Z, and compression X, between two 



220 ENGINEERING THERMODYNAMICS 

pressures, and when the law of expansion is PV S = const, and s anything except 
unity, but constant. 

The mean effective pressure, pounds per square inch, is obtained by divid- 
ing the expression for work by 144Z), giving 



, (282) 



(m.c.p.) = (in.pr.) j Z+^j[l - (5)" ( mean for>d 



-(bk.pr.) ( (l_JE)+gK(*?y Vi] 1 (mean bk.pr.) 
[ s ~ L L \ c / J J 



which is the general expression for mean effective pressure for this cycle. 

It was pointed out in Section (5) that the cubic feet of fluid admitted at the 
initial pressure was not represented by AB, Fig. 60, but by QB, and the same 
is true for this case, so that the 



But when the expansion and compression laws have the form PV s = 



Whence 



^-y ..... (283) 



Eq. (283) gives for these cases what was given by Eq. (263) for the logarithmic 
law, the cubic feet of fluid supplied at the initial pressure for the displacement 
of D cubic feet in terms of cut-off, clearance, compression and the pressures. 
From this, by division there is found the volume of high pressure fluid per 
cu.ft. of displacement 

(Sup.Vol.) 
D 

The consumption is given by the general expression already derived, 
Eq. (34), from which is obtained, 

Consumption Ibs. per hr. per I.H.P. 



WORK OF PISTON ENGINES 221 

Eq. (285) gives the water rate or gas consumption in terms of mean effective 
pressure, initial and back pressure, cut-off, clearance, compression and initial 
fluid density. Introducing the density at the back pressure by the relation, 

PIT _ p V s 
a V Q * e r e ) 



J^ 

jj = Si /in.pr.X * 
V, 8 2 Vbk.pr./ 

i.pr.AT 
k^rV ' 



in. 
bk. 



[there results 



Consumption Ibs. per hr. per I.H.P. = ^J 5 \(Z+c) 81 - (X+c) 8 2 1 , . 



(286) 



rhich is identical with Eq. (268) and is, as previously observed, a general expres- 
sion, no matter what the laws of expansion and compression, in terms of 
mean effective pressure, cut-off, clearance, compression and the initial and 
final steam density. 

The first special case of full admission, no compression might at first thought 
appear to be the same as in the preceding section, where the logarithmic law 
was assumed to hold, and so it is as regards work and mean effective pressure, 
Eq. (269) and (270), but referring to Fig. 61 it will be seen that since the expo- 
nential law is now assumed instead of the logarithmic, the point Q will be dif- 
ferently located (nearer to A than it was previously if s is greater than 1), and 
hence the supply volume QB is changed, and its new value is 



in.pr. 



(287) 



Hence, consumption stated in terms of initial density of the fluid 81, is 



Consumption Ibs. per hr. per I.H.P. 

13 ' 75 



(in.pr.)-(bk.pr.) 



lie 



. . (288) 



Stated in terms of initial and final densities, the expression is as before, Eq. (273). 
The second special case, complete expansion and compression is again repre- 
sented by Fig. 62. From the law of expansion it is evident that 



222 ENGINEERING THERMODYNAMICS 

or stated in symbolic form, 



whence 



Again referring to Fig. 62, 

'^ ?Y -cD 



V c -Va 
Whence 



-y re 'a ywt.pt. 

-A = ^~ 



(290) 



Eq. (289) gives the cut-off as a fraction of displacement necessary to give 
complete expansion, and (290), the compression fraction to give complete com- 
pression, both in terms of clearance, initial and back pressures, and the exponent 
s, in the equation of the expansion or compression line, PV S const. 

The work of the cycle becomes for this special case, by substitution in Eq. 
(281), 

J^ s-1 

F=144Z)(in.pr.) 



and the mean effective pressure, Ibs. per sq.in., is 

( 292 ) 



hence 



The volume of fluid supplied is, 

(Sup.Vol.)=ZD, .... .... (293) 



Consumption, Ibs. per hr. per I.H.P.= 

m.e.p 



but 
Z 



8-1 

r ^ s /bk.prA s 

(m.pr.) 1- - -^~ 

y s-l[ \in.pr. / J, 



WOEK OF PISTON ENGINES 223 

whence 

Consumption Ibs. fluid per hr. per I.H.P. is, 

13,750XSi 



s-l 



(294) 



,. , s /bk.pr.X "I 

(m.pr.) 1- - 

s-lL \in.pr. / 



the expression for smallest consumption (or water rate if steam) of fluid for 
the most economical hypothetical cycle, which may it be noticed, is again in- 
dependent of clearance. 

The expressions for work and mean effective pressure are not, however, 
independent of clearance, and hence, according to the hypothetical cycles here 
considered, it is proved that large clearance decreases the work capacity of a 
a cylinder of given size, bjit does not affect the economy, provided complete 
expansion and compression are attained, a conclusion similar to that in regard 
to clearance effect on compressor capacity and economy. Whether the actual 
performance of gas or steam engines agrees with this conclusion based only 
on hypothetical reasoning, will be discussed later. 

Example 1. What will be the horse-power of and steam used per hour by the 
following engine: 12xl8-in. double-acting, 200 R.P.M., 7 per cent clearance, initial 
pressure 85 Ibs. per square inch gage, back pressure 15 Ibs. per square inch absolute, 
and expansion such that s = 1.3. 

(a) cut-off =50 per cent; compression = 30 per cent; 

(b) complete expansion and compression; 

(c) no expansion or compression; 

NOTE: 8 for 85 Ibs. gage =.18; for 15 Ibs. absolute. = .03. 
(a) From Eq. (282) 



59.8X1.5X113.1X400 
- 



From Eq. (286) 
Steam per hour per I.H.P. 



, - X.18-(.37) x.03] =20.9 Ibs. 
(m.e.p.) |_ j 5 

/. Steam per hour = 122 X20.9 = 2560 Ibs. 



224 ENGINEERING THERMODYNAMICS 

(6) From Eq. (292) 

s f x /bk.pr.\r "ir /bk.pr.\~~ 
(m.e. P = (in.pr.)-- I [(l +C )(--y ^Jl^ta^ J' 



s-l 



I .in., 



26.2X1.5X113.1X200 
3000 



From Eq. (294) steam used I.H.P. per hour is, 

_ 13,7508i _ 13,750 X.18 _ ==16>51bs . ) 

T /bk.prAT 1 ] 1.3 I" /ISXTT] 

(-p^r-TL^tpTrj J ^"T'ViooJ J 



hence total steam per hour = 16.5 X64 = 1060 Ibs. 

(c) From Eq. (270) which holds for any value of s, m.e.p. =100-15 =85 Ibs. sq.in. 






From Eq. (288) steam per I.H.P. hour 

1 =24 5 lbs 



(m.e.p.) L \in.pr. / J 85 

and total steam per hour = 174.5X24.5 =2475 lbs. 

Prob. 1. Two simple engines, each 12 Xl8 ins., with 6 per cent clearance are operated 
at ^ cut-off and with 20 per cent compression. One is supplied with air at 80 lbs. 
per square inch gage, and exhausts it to atmosphere; the other with initially dry steam 
which becomes wet on expansion and which also is exhausted to atmosphere. For a speed 
of 200 R.P.M. what is the horse-power of each engine and the cubic feet of stuff supplied 
per horse-power hour ? 

Prob. 2. A crank-and-flywheel two-cylinder, double-acting, pumping engine is 
supplied with dry steam and the expansion is such that it remains dry until exhaust. 
The cylinder size is 24x36 ins., cut-off to give perfect expansion, clearance 5 per cent, 
compression to give perfect compression, initial pressure 50 lbs. per square inch ab- 
solute. back pressure 5 lbs. per square inch absolute. What is the horse-power and 
water rate? What would be the horse-power and water rate of a full-stroke pump of the 
same size and clearance but having no compression, running on the same pressure range 
and quality of steam. 

NOTE: 8 for 50 lbs. absolute = .12, 8 for 15 lbs. absolute = .0387. 

Prob. 3. Should the cylinder of the following engine be so provided that th< 
steam was always kept dry, would there be any change in the horse-power developed a: 



WORK OF PISTON ENGINES 225 

compared with steam expanded adiabatically, and how much? Cylinder 20x24 ins., 
initial pressure 125 Ibs. per sq. in. gage, back pressure 26 ins. vacuum, standard barom- 
eter, clearance 6 per cent, cut-off f, compression 10 percent, and speed 125 R.P.M. 

Prob. 4. What will be the total'steam used per hour by a 20x28-in. double-acting 
engine running at 150 R.P.M. if the initial pressure be 125 Ibs. per square inch absolute, 
back pressure one atmosphere, clearance 8 per cent, compression zero, for.cut-off 
I, i, f, and %, if steam expands adiabatically and is originally dry and saturated? 

NOTE: 8 for 125 Ibs. absolute = .283, 8 for 15 Ibs. absolute = .0387. 

Prob. 5. An engine which is supplied with superheated steam is said to have an 
indicated water rate of 15 Ibs. at | cut-off and one of 25 Ibs. at | cut-off. See if 
this is reason able for the folio wing conditions: engine is 15x22 ins., 7 per cent clearance, 
no compression, initial pressure 100 Ibs. per square inch gage., back pressure 28-in. 
vacuum, barometer 30 ins. and speed 180 R.P.M. 

NOTE: 8 for 100 Ibs. gage =.262, 8 for 28 in. Hg = .0029. 

Prob. 6. The tank capacity of a locomotive is 4500 gals. The cylinders are 
18X36 ins., initial pressure 200 Ibs. per square inch gage, exhaust atmospheric, cut-off 5, 
clearance 7 per cent, speed 200 R.P.M., no compression. The stream is dry at 
start and expansion adiabatic, how long will the water last if 40% condenses during 
admission? 

NOTE: 8 for 200 Ibs. gage = .471, 8 for 15 Ibs. absolute = .0387. 

Prob. 7. To drive a hoist, an air engine is used, the air being supplied for ^ 
stroke at 80 Ibs. per squjfc&inch gage expanded adiabatically and exhausted to atmos- 
sphere. If the clearance is Sjf per cent and there is no compression how many cubic 
feet of air per hour per horse-power will be needed? What, with complete compression? 

Prob. 8. A manufacturer rates his 44x42-in. double-acting engine with a speed 
of 100 R.P.M. at 1000 H.P. when running non-condensing, initial pressure 70 Ibs. 
per square inch gage and cut-off i. No clearance is mentioned and nothing said about 
manner of expansion. Assuming s = 1.0646, find on what clearance basis this rating is 
made. 

Prob. 9. The water supply of a town is supplied by a direct-acting non-condensing 
pump with two cylinders, each 24x42 ins., with 10 per cent clearance, and no com- 
pression, initial pressure being 100 Ibs. per square inch gage. What must be the size of the 
fcteam cylinder of a crank-and-flywheel pump with 6 per cent clearance to give the same 
jhorse-power on the same steam and exhaust pressures with a cut-off of ? Speed in 
.each case to be 50 R.P.M. 

NOTE: 8 for 100 Ibs. gage =.262, 8 for 15 Ibs. abs.=.0387. 

7. Action of Fluid in Multiple-expansion Cylinders. General Description 

Structure and Processes. When steam, compressed air, or any other high 
jressure working fluid is caused to pass through more than one cylinder in 
jries, so that the exhaust from the one is the supply for the next, the engine 
5, in general, a multiple-expansion engine, or more specifically, a compound 
n the operations are in two expansion stages, triple for three, and quadruple 
For four stages. It must be understood that while a compound engine is one 

which the whole pressure-volume change from initial to back pressure takes 
>lace in two stages, it does not necessarily follow that the machine is a two- 
jylinder one, for the second stage of expansion may take place in two cylinders, 

each of which, half of the steam is put through identical operations; this 



226 ENGINEERING THERMODYNAMICS 

would make a three-cylinder compound. Similarly, triple-expansion engines, 
while they cannot have less than three may have four or five or six cylinders 
Multiple expansions engine, most of which are compound, are of two classes 
with respect to the treatment and pressure-volume changes of the steam, first 
wthout receiver, and second, with receiver. A receiver is primarily a chamber 
large in proportion to cylinder volumes, placed between the high- and low-pres- 
sure cylinders of compounds or between any pair of cylinders in triple or 
quadruple engines, and its purpose is to provide a reservoir of fluid so that 
the exhaust from the higher into it, or the admission to the lower from it, will 
be accomplished without a material change of pressure, and this will be accom- 
plished as its volume is large in proportion to the charge of steam received 
by it or delivered from it. With a receiver of infinite size the exhaust line of 
a high-pressure cylinder discharging into it will be a constant-pressure line, 
as will also the admission line of the low-pressure cylinder. When, however, 
the receiver is of finite size high-pressure exhaust is equivalent to increasing 
the quantity of fluid in the receiver of fixed volume and must be accompanied 
by a rise of pressure except when a low-pressure cylinder may happen to be 
taking out fluid at the same rate and at the same time, which in practice never 
happens. As the receiver becomes smaller in proportion to the cylinders, the 
pressure in it will rise and fall more for each high-pressure exhaust and low- 
pressure admission with, of course, a constant average value. The greatest 
possible change of pressure during high-pressure exhaust and low-pressure admis- 
sion would occur when the receiver is of zero size, that is when there is none at all, 
in which case, of course, the high-and low-pressure pistons must have synchronous 
movement, both starting and stopping at the same time, but moving either in 
the same or opposite directions. When the pistons of the no-receiver compound 
engines move in the same direction at the same time, one end of the high- 
pressure cylinder must exhaust into the opposite end of the low; but with 
oppositely moving pistons, the exhaust from high will enter the same end of the 
low. It is plain that a real receiver bf zero volume is impossible, as the connect- 
ing ports must have some volume and likewise that an infinite receiver is equally 
impracticable, so that any multiple-expansion real engine will have receivers 
of finite volume with corresponding pressure changes during the period when a 
receiver is in communication with a cylinder. The amount of these pressure 
changes will depend partly on the size of the receiver with respect to the cylin- 
ders, but also as well, on the relation between the periods of flow into receiver, 
by high-pressure exhaust and out of it, by low-pressure admission, which 
latter factor will be fixed largely by crank angles, and partly by the settings 
of the two valves, relations which are often extremely complicated. 

For the purpose of analysis it is desirable to treat the two limiting cases 
of no receiver and infinite receiver, because they yield formulas simple enough to 
be useful, while an exact simple solution of the general case is impossible. These 
simple expressions for hypothetical cases which are very valuable for estimates 
and approximations are generally close to truth for an actual engine especially 
if intelligently selected and used. 



WORK OF PISTON ENGINES 227 

Receivers of steam engines may be simple tanks or temporary storage 
hambers or be fitted with coils or tubes to which live or high-pressure steam 
s supplied and which may heat up the lower pressure, partly expanded steam 
mssing from cylinder to cylinder through the receiver. Such receivers are 
cheating receivers, 'and as noted, may heat up the engine steam or may evapo- 
ate any moisture it might contain. As a matter of fact there can be no 
leating of the steahi before all moisture is first evaporated, from which 
t appears that the action of such reheating receivers may be, and is quite 
iomplicated thermally, and a study of these conditions must be postponed 
ill a thermal method of analysis is established. This will introduce no 
serious difficulty, as such reheating receivers assist the thermal economy 
)f the whole system but little and have little effect on engine power, likewise 
ire now little used. Reheating of air or other gases, as well as preheating 
'hem before admission to the high-pressure cylinder is a necessary practice, 
when the supply pressure is high, 'to prevent freezing of moisture by the gases, 
which get very cold in expansion if it be carried far. This is likewise, however, 
a thermal problem, not to be taken up till later. 

Multiple-expansion engines are built for greater economy than is possible 
n simple engines and the reasons are divisible into two classes, first mechanical, 
and second thermal. It has already been shown that by expansion, work is 
obtained in greater amounts as the expansion is greater, provided, of course, 
xpansion below the back pressure is avoided, and as high initial and low 
Dack pressures permit essentially of most expansion, engines must be built 
capable of utilizing all that the steam or compressed gas may yield. If 
ateam followed the logarithmic law of expansion, pressure falling inversely 
with volume increase, then steam of 150 Ibs. per square inch absolute 
expanding to 1 Ib. per square inch absolute would require enough ultimate 
cylinder space to allow whatever volume of steam was admitted up to cut-off 
to increase 150 times. This would involve a valve gear and cylinder structure 
capable of admitting T J-<j = .0067 of the cylinder volume. It is practically 
impossible to construct a valve that will accurately open and close in this 
necessarily short equivalent portion of the stroke. This, however, is not the 
worst handicap even mechanically, because actual cylinders cannot be made 
without some clearance, usually more than 2 per cent of the displacement and 
n order that any steam might be admitted at all, the clearance in the example 
would have to be less than .67 per cent of the total volume, which is quite 
impossible. These two mechanical or structural limitations, that of admission 
valve gear and that of clearance limits, supply the first argument for multiple- 
expansion engines, the structure of which is capable of utilizing any amount 
of expansion that high boiler pressure and good condenser vacuum make 
available. For, if neglecting clearance, the low-pressure cylinder had ten 
iimes the volume of the high, then the full stroke admission of steam to the high 
followed by expansion in the low would give ten expansions, while admission 
k) the high for ^ of its stroke would give 15 expansions in it, after which this 
inal volume would increase in the low ten times, that is, to 150 times the original 



22S ENGINEERING THERMODYNAMICS 

volume, and cylinder admission of ^ or 6.7 per cent is possible with ordinary 
valve gears, as is also an initial volume of 6.7 per cent of a total cylinder 
volume, even with clearance which in reasonably large engines may be noi 
over 2 per cent of the whole cylinder volume. 

It is evident that the higher the initial and the lower 'the back pressure* 
the greater the expansion ratio will be for complete expansion, and as in steair 
practice boiler pressures of 225 Ibs. per square inch gage or approximately 24( 
Ibs. per square inch absolute with vacuum back pressures as low as one or ever 
half a pound per square inch are in use, it should be possible whether desirable 
or not, to expand to a final volume from 250 to 500 times the initial in round num- 
bers. This is, of course, quite impossible in simple engine cylinders, and as i1 
easy with multiple expansion there is supplied another mechanical argument foi 
staging. Sufficient expansion for practical purposes in locomotives and lane 
engines under the usually variable load of industrial service is available for ever 
these high pressures by compounding, but when the loads are about constant 
as in waterworks pumping engines, and marine engines for ship propulsion 
triple expansion is used for pressures in excess of about 180 Ibs. gage. 

Use of very high initial and very low back pressures will result in simple 
engines, in a possibility of great unbalanced forces on a piston, its rods, pirn 
and crank, when acting on opposite sides, and a considerable fluctuation in tan- 
gential turning force at the crank pin. Compounding will always reduce the 
unbalanced force on a piston, and when carried out in cylinders each of whict 
has a separate crank, permits of a very considerable improvement of turning 
effort. So that, not only does multiple expansion make it possible to utilize 
to the fullest extent the whole range of high initial and low back pressures 
but it may result in a better force distribution in the engine, avoiding shocks 
making unnecessary, excessively strong pistons, and rods and equalizing turn- 
ing effort so that the maximum and minimum tangential force do not deparl 
too much from the mean. 

The second or thermal reason for bothering with multiple-expansion com- 
plications in the interest of steam economy is concerned with the preventior 
of steam loss by condensation and leakage. It does not need any elaborate 
analysis to show that low-pressure steam will be cooler than high-pressure 
steam and that expanding steam in a cylinder has a tendency to cool the 
cylinder and piston walls, certainly the inner skin at least, so that aftei 
expansion and exhaust they will be cooler than after admission; but as 
admission follows exhaust hot live steam will come into contact with coo' 
walls and some will necessarily condense, the amount being smaller the less 
the original expansion; hence in any one cylinder of a multiple-expansior 
engine the condensation may be less than a simple engine with the same range 
of steam pressures and temperatures. Whether all the steam condensatior 
during admission added together will equal that of the simple engine or no1 
i^ another .question. There is no doubt, however, that as the multiple expan- 
sion engine can expand usefully to greater degree than a simple engine, anc 
so cause a lower temperature by expansion, that it has a greater chance tc 



WORK OF PISTON ENGINES 229 

eevaporate some of the water of initial condensation and so get some work 
ut of the extra steam so evaporated, which in the simple engine might have 
emained as water, incapable of working until exhaust opened and lowered 
he pressure, when, of course, it could do no good. It is also clear that steam 
>r compressed-air leakage in a simple engine is a direct loss, whereas in a 
:ompound high-pressure cylinder leakage has at least a chance to do some 
vork in the low-pressure cylinder. The exact analysis of the thermal reasons 
or greater economy is complicated and is largely concerned with a study of 
;team condensation and reevaporation, but the fact is, that multiple-expan- 
sion engines are capable of greater economy than simple. The thermal analysis 
nust also consider the influence of the reheating receiver, the steam-jacketed 
vorking cylinder, and the use of superheated steam, their effects on the pos- 
>ible work per pound of steam and the corresponding quantity of heat expended 
;o secure it, and for air and compressed gas the parallel treatment of pre- 
leating and reheating. 

To illustrate the action of steam in multiple-expansion engines some indi- 
3ator cards are given for a few typical cases in Figs. 63 to 66, together 
with the combined diagrams of pressure-volume changes of the fluid in all 
3ylinders to the same scale of pressures and volumes, which, of course, makes 
ihe diagram look quite different, as indicator cards are usually taken to the 
same base length, fixed by the reducing motion, and to different pressure 
Scales, to get as large a height of diagram as the . paper will permit. Fig. 63 
bows four sets of cards taken from an engine of the compound no-receiver 
jype, namely, a Vauclain compound locomotive. In this machine there are 
f\vo cylinders, one high pressure and one low, on each side, the steam from 
he high pressure exhausting directly into the low-pressure cylinder so that 
;the only receiver space is made up of the clearance and connecting passages 
I between the cylinders. Starting with set A, the cards show a decreasing 
|igh pressure cut-off of 76 per cent in the case of set A to 54 per cent in the 
iase of set D. The letters A, B, C and D refer in each case to admission, cut- 
, release and compression, the use of primed letters denoting the low-pressure 
cylinder. 

In set A the high-pressure admission line AFB may be considered as made 
lp of two parts, the part AF representing pressure rise at constant volume, 
wh ; ch is the admission of steam to the clearance space at dead center to raise the 
"essure from that at the end of compression to that of boiler pressure. From 
m to B admission occurred at constant pressure, steam filling the cylinder 
lolume as the piston moved outward. At B cut-off or closure of the steam 
jalve occurred and the steam in the cylinder expanded. At C, release or open- 
4f the exhaust valve of the high-pressure cylinder occurred and the admission 
Valve of the low-pressure cylinder opened, the steam dropping in pressure until 
the pressure in both high- and low-pressure clearance became equal, and then 
mpanding in both cylinders, as the exhaust from the high and admission to 
j the low occurred, the exhaust line CD of the high pressure and the admission 
:line F'B' of the low pressure being identical except for the slight pressure drop 



230 



ENGINEERING THERMODYNAMICS 






(A) 




(B) 







(D) 

FIG. 63. Set of Indicator Cards from Vauclain Locomotive Illustrating the No-receiver 

Compound Steam Engine. 



WOKK OF PISTON ENGINES 231 

in the passages between the high- and the low-pressure cylinders. At D the 
high-pressure exhaust valve closed and compression of the steam trapped in 
the high-pressure cylinder occurred to point A, thus closing the cycle. From 
point D' in the low-pressure cylinder, which corresponds to D in the high 
pressure, no more steam was admitted to the low-pressure cylinder. What 
steam there was in the low expanded to the point C" when the exhaust valve 
opened and the pressure dropped to the back pressure and the steam was 
exhausted at nearly constant back pressure to D', when the exhaust valve closed 
and the steam trapped in the cylinder was compressed to A', at which point 
steam was again admitted and the cycle repeated. 

In set B the cycle of operation is exactly the same as in set A. In set C 
the cycle is the same as in A, but there are one or two points to be especially 
noted, as they are not present in set A. The admission line of the high-pres- 
sure cylinder is not a constant pressure, but rather a falling pressure one, due to 
throttling of the steam, or "wire drawing," as it is called, through the throttle 
valve or steam ports, due to the higher speed at which this card was taken. 
It will also be noticed that the compression pressure is higher in this case, due 
to earlier closing of the exhaust valve, which becomes necessary with the type of 
valve gear used, as the cut-off is made earlier. In the low-pressure card it will 
be seen that the compression pressure is greater than the admission pressure 
and hence there is a pressure drop instead of rise on admission. In set D the 
peculiarities of C are still more apparent, the compression in high-pressure 
cylinder being equal to admission pressure and above it in the low-pressure 
cylinder. The wire drawing is also more marked, as the speed was still 
higher when this set of cards was taken. 

In Fig. 64, one set of the cards of Fig. 63 is redrawn on cross-section 
paper and then combined. Cards taken from the different cylinders of a 
multiple-expansion engine will in nearly all cases have the same length, the great- 
est that can be conveniently handled by the indicator, and will be to two different 
pressures scales, in as much as that indicator spring will be chosen for each 
cylinder which will give the greatest height of card consistent with safety to 
- the instrument. To properly compare the cards they must be reduced to the 
I same pressure scale, and also to the same volume scale. As the lengths 
I represent volumes, the ratio of the two volume scales will be as that of the 
j cylinder volumes or diameters squared. Hence, the length of the high-pressure 
? card must be decreased in this ratio or the low increased. As a rule it is found 

t| more convenient to employ the former method. When the cards have been 
reduced to a proper scale of pressures and volumes the clearance must be 
> added to each in order that the true volume of the fluid may be shown. 
I The cards may now be placed with these atmospheric lines and zero volume 
I lines coinciding and will then appear in their true relation. In this case the 
cylinder ratio was 1.65, the indicator springs 100 Ibs. and 70 Ibs. respectively 
and clearance 5 per cent in each cylinder. 

The steps in combining the cards were as follows: The zero volume lines were 
first drawn perpendicular to the atmospheric line and at a distance from the end of 



232 



ENGINEERING THERMODYNAMICS 



the card equal to the length of the card times the clearance. PV axes were laid 
off and a line drawn parallel to the zero-pressure line at a distance above it equal to 
14.7 Ibs. to scale of combined diagram. This scale was taken to be that of the 
high-pressure diagram. A number of points A'B'C', etc., were then chosen 
on the low-pressure card, and the corresponding points a' b'c', etc., plotted by 
making the distances of a', &', etc., from the zero-volume line equal to thoce 
of A',B', etc., and the distances of the new points above the atmosphere .7 
the distances of the original. By joining the points as plotted, the new diagram 
for the low-pressure card was formed. The high-pressure card was then redrawn 



c 



A' 



FIG. 64. Diagram to Show Method of Combining the High- and Low-pressure Cylinder 
Indicator Cards of the No-receiver Compound Engine. 

by taking a number of points A, B, C, etc., and plotting new points a, b, c, 

etc., so that the distances of a, 6, c, etc., from the zero-volume line were the 

l.bo 

distances of A, B, C, etc., while the distances of new points above the atmos- 
pheric line were the same as for the original points. 

In Fig. 65 are shown two cards from a compound steam engine with receiver. 
Diagram A shows the cards as taken, but transferred to cross-section paper for 
ease in combining, and with the zero-volume axis added. On the high-pressure 
card admission occurred practically at constant volume, piston being at rest 
at dead center, at A, bringing the pressures in the cylinder up to the initial 
pressure at B. Admission continued from B to C at nearly constant pressure, 
the piston moving slowly with correspondingly small demand for steam and 
consequently little wire drawing. From C to D the piston is moving more 
rapidly and there is in consequence more wire drawing, admission being no 
longer at constant pressure. At D the steam valve closes and expansion occurs, 
to E y whefe release occurs, the pressure falling to that in the receiver. From 
F to G exhaust occurs with increase of pressure due to -the steam being forced 
into the receiver, (receiver-}- decreasing H.P. cyl.vol.) while from G to H the 



WORK OF PISTON ENGINES 



233 



pressure falls, due to the low-pressure cylinder taking steam from the receiver 
and consequently volume of receiver, (receiver + increasing L.P. cyl.vol.+ 
decreasing H.P. cyl.vol.) increasing. At H exhaust closes, a very slight com- 
pression occurring from H to A . 

On the low-pressure card, admission occurred at A' and continued to B' at 
constant volume, the piston being on dead center as from A to B in high-pressure 
cylinder. From B' to C" admission occurred with falling pressure due to increase 
in receiver volume, (receiver -f increasing L.P. cyl.vol.), and from C' to D' 
admission still took place, but with less rapidly falling pressure, as high-pres- 
sure cylinder is now exhausting and receiver volume, (receiver + increasing 
L.P. cyl. vol. + decreasing H.P. cyl.vol.) was receiving some steam as well as 




! FIG. 65. Indicator Cards from a Compound Engine with a Receiver, as Taken and as 

Combined. 

Delivering. At D' admission ceased and expansion took place to E' where 
felease occurred, the pressure falling to the back pressure and continuing from 
F' to G f , where the exhaust valve closed and compression took place to A', 
thus completing the cycle. At H f leakage past the exhaust valve was so great 
as to cause the curve to fall off considerably from H' to A', instead of con- 
mnuing to be a true compression curve, ending at /, as it should have done. 
iffhe combined diagrams are shown in B. 

In Fig. 66 are shown a set of three cards from a triple-expansion pumping 
ingine with large receivers and cranks at 120. In diagram A the cards are 
Hiown with the same length and with different pressure scales as taken, but 
with the zero volume line added and transferred to cross-section paper. On 
the high-pressure card admission occurred at A, causing a constant volume 
ressure rise to B, the piston being at rest with the crank at dead center. From 
P to C admission occurred at nearly constant pressure to C, where steam was 



234 



ENGINEERING THERMODYNAMICS 



cut off and expansion took place to D. At this point release occurred, the 
pressure dropping at constant volume to E with the piston at rest. From E 
exhaust took place with slightly increasing pressure, since the intermediate 
cylinder was taking no steam, the intermediate piston being beyond the point 
of cut-off. The pressure rise is slight, however, due to the size of the receiver, 
which is large compared to the cylinder. At two-thirds of the exhaust stroke, 
point F, the back pressure became constant and then decreased, for at this 
point the speed of the intermediate piston increased and the receiver pressure 
fell. At G exhaust closed and a slight pressure rise occurred to A, due to the 
restricted passage of the closing exhaust valve. On the intermediate card 



At H 



B" 



Atn 



Atm 



FIG. 66. Indicator Cards from a Triple-expansion Engine with Receiver as Taken and 

Combined. 

admission occurred at A', the pressure rising to B'. From B f the admissio 
was at nearly constant pressure to X while the piston speed was low and the 
at a falling pressure to C". Pressure was falling, since the steam was supplie 
from a finite receiver into which no steam was flowing during intermediati 
admission. At C" cut-off occurred and steam expanded to D', where relea:| 
took place, and the steam was exhausted. As in the case of the high-pressu j 
cylinder the back pressure was rising for two-thirds of the stroke, since tl< 
steam was being compressed into the receiver or rather into a volume ma<( 
up of receiver and intermediate cylinder volume, which is, of course, a decreoj- 
ing one, since the cylinder volume is decreasing. At two-thirds of the stro h 
the low-pressure cylinder begins to take steam and the receiver volume I 

i 



WOEK OF PISTON ENGINES 235 

now increased, inasmuch as it was made up of the receiver portion of the inter- 
mediate cylinder and a portion of the low-pressure cylinder, and the low- 
pressure cylinder volume increased faster than intermediate decreased for the 
same amount of piston travel. At G' exhaust closed and a slight compression 
occurred to A', thus completing the cycle. 

On the low-pressure card admission occurred at A" and the pressure rose at 
constant volume to B", and then admission continued first at constant pressure 
and then falling, as in the intermediate cylinder, to the point of cut-off 
at C" '. From here expansion took place to D n '. At this point the exhaust 
valve opened, the pressure fell nearly to back pressure at E", and the steam 
was exhausted at practically constant back pressure to G" ', where the exhaust 
valve closed and there was compression to A", thus completing the cycle. 
The combined diagram is shown in B. 






Prob. 1. In Fig. 67 are shown six sets. of indicator cards from compound engines. 
The cylinder sizes and clearances are given below. Explain the cylinder events and the 
shape of lines for each card and form a combined diagram for each set. 

No. 1. From a four- valve Corliss engine, 26x48 ins., with 3 per cent clearance in 
each cylinder. 

No. 2. From a single- valve engine, 12x20x12 ins., with 33 per cent clearance in 
high-pressure cylinder and 9 per cent in low. 

No. 3. From a four-valve Corliss engine 22x44x60 ins., with 2 per cent clearance 
in the high-pressure cylinder and 6 per cent in low. 

No. 4. From a single- valve engine 18 X30X16 ins., with 30 per cent clearance in the 
high-pressure cylinder and 8 per cent in the low. 

No. 5. From a single- valve engine 11^X18^X13 ins., with 7 per cent clearance in 
the high and 10 per cent in the low. 

No. 6. From a double- valve engine 14 X 28 X2^ ins.,, with 3.5 per cent clearance in 
the high-pressure cylinder and 6.5 per cent in the low. 

Prob. 2. In Fig. 68 are shown four sets of indicator cards from triple-expansion 
marine engines. The cylinder sizes and clearances are given below. Explain the cylinder 
events and the shape of the lines of each card and form a combined diagram of each set. 

No. 1. From the engine of a steam-ship, cylinders 21.9x34x57 ins. X39 ins. with 
6 per cent clearance in each and fitted with simple slide valves. 

No, 2. From an engine 20x30x50x48 ins. 

No. 3. From the engine of a steam-ship with cylinders 22x35x58x42 ins., 
assume clearance 7 per cent in each cylinder. 

No. 4. From the engine of the steamer "Aberdeen, " with cylinders 30 X45 X70 X54 
ins., and with 4 per cent clearance in the high, 7 per cent in the intermediate, and 8 per 
jj&ent in the low. 

Prob. 3. In Fig. 69 are shown some combined cards from compound engines. 
Explain the cylinder events and the shape of the lines and reproduce the indicator cards. 

Prob. 4. In Fig. 70 are ahown some combined cards from triple-expansion engines. 
'Draw the individual cards and explain the cylinder events and shape of lines! 

8. Standard Reference Cycles or PV Diagrams for the Work of Expansive 
Fluids in Two-cylinder Compound Engines. The possible combinations of 
jadmission with all degrees of expansion for forward strokes and of exhaust with 



236 ENGINEERING THERMODYNAMICS 

all degrees of compression for back strokes, with and without clearance, in 
each of the two cylinders of the compound engine, that may have any volume 
relation one to the other and any size of receiver between., and finally, any sort 
of periodicity of receiver receipt and discharge of fluid, all make possible a 

-100 



GO 



-20 
-0 




No. 2 



lO-i 

o- 

10- 



No. 3 




No. 4 




-40 
-20 



-0 
-10 



-40 
-20 

-0 
-JO 





-100 



-60 



-20 
-0 




120 



-0 



No. 6 





10 
-0 



FIG. 67. Six Sets of Compound Engine Indicator Cards. 



very large number of cycles. In order that analysis of these conditions of work- 
ing may be kept within reasonable space, it is necessary to proceed as was 
done with compressors and simple engines, concentrating attention on sucl 
type forms as yield readily to analytical treatment and for which the formula; 



WORK OF PISTON ENGINES 



are simple even if only approximate with respect to actual engines, but, of 
course, keeping in mind the possible value of the formulas, as those that teach 




FIG. 68. Four Sets of Triple-expansion Engine Indicator Cards. 

no principles or fail to assist in solving problems must be discarded as useless. 
The work that fluids under pressure can do by losing that pressure is no 



238 



ENGINEERING THERMODYNAMICS 



different in compound than in simple engines, if the fluid has a chance to do 
what ii can. Provided the structure is such as will not interfere with the com- 
pleteness of the expansion, and no fluid is wasted in filling dead spaces without 



130-1 



200- 



100- 




100- 




No. 7 





No. 8 




Fio. 69. Combined Diagrams of Compound Engines. 

working, then the work per cubic foot or per pound of fluid is the same for simple 
compound and triple engines. Furthermore, there is a horse-power equiv- 
alence between the simple and rorn pound, if, in the latter case the steanr 



\\oicu OK PISTON I:\<JI\KS 



Up t. ruf 





no 



,-. 



100 



r 



onlv in flu I mr 

'\,-|l :>: l( <i \ < :r :i 





X.. 1 



N.. ft 



l-'i,, ,0 ( 'uinhln.-.l I,,,,MM.M.. ..i I,,,, I, , t|uiiiMnii I'',..,.,....' 

il ..II IIH'MII ... in.lrt TliiM flu.ill.l lir rl.'MI 1 1 <Uli H ruiii| 1:1 1 \^ t 71, 

MM, I II III l-ii- ,11. i, |.i. . IllillfJ lll< . II --I ||, 11,,,. I. , ur.n, VN ill,.. ill 



240 



ENGINEERING THERMODYNAMICS 



clearance and with complete expansion, the volume admitted, AB, expand 
to the back pressure on reaching the full cylinder volume DC, and exhausl 
at constant back pressure, the work represented by the area A BCD. It shoul 
be clear that no difference will result in the work done if a line be drawn acros 
the work area as in Fig. 7 IB, all work done above the line HG to be develope 
in the high-pressure cylinder and that belowjn the low. This is merely equn 
ulent to saying that a volume of steam AB is admitted to the high-pressui 
cylinder expands completely to the pressure at G on reaching the full higl 
pressure cylinder volume, after which it exhausts at constant pressure (int 
a receiver of infinite capacity), this same amount being subsequently admitte 
without change of pressure to the low-pressure cylinder, when it again expanc 
completely. Thus, it appears that the working of steam or compressed a 



Vol. Admitted to Cylinder 



\ 



Displacement Volume 



I .Ad: 



Vol. Admitted to H 



H.P. Displ icement 



B 



Volume Admitted to L.P. Cyl. 



Cyl rider 



L.P. Displacement - 



FIG. 71. Diagram to Show Equality of Work for Expansion in One-cylinder Simple and 
Two-cylinder Compound Engines for the Same Rate of Expansion. 

in two successive cylinders instead of one will in no way change the maximu 
amount of work a cubic foot supplied can do, the compounding merely maki 
it easier to get this maximum. In simple engine cases, Fig. 71A, the cut-off in p 

AB 

cent of stroke is 100 X 7^, which is a very small value, leaving but little time 
.DC 

open and close the admission valve, whereas in the compound case the per ce 
cut-off in the high-pressure cylinder is lOOXTTT^and in the low-pressure cylind 



100X ,^, which are quite large enough ratios to be easily managed with ordina 

L/L> 

valve gc.-i 

Compounding docs, however, introduce possibilities of loss not present 
the sinjilc-slugc expansion, if the dimensions or adjustments are not rig 
which may be classed somewhat improperly as receiver losses, and these 



WORK OF PISTON ENGINES 



241 



of two kinds, one due to incomplete 
expansion in the high and the other 
to over-expansion there. Thus, in 
Fig. 72, if ABCEFGDA represent a 
combined compound diagram for 
the case of complete expansion in 
the high-pressure cylinder continued 
in the low without interruption but 
incomplete there, DC represents the 
volume in the low-pressure cylinder 
at cut-off, and at the same time the 
total high-pressure cylinder volume. 

If now, the low-pressure cut-off 
be made to occur later, Fig. 73, 
then the volume that the steam _ 
would occdpy when expansion began F 7 
in the low-pressure cylinder is rep- 
resented by D'C'. This adjust- 
ment could not, of course, change the 



\ 



\ 



WIG. 73. Diagram to Show Effect of Lengthening 
[ L.P. Cut-off Introducing a Receiver Loss Due to 
? Incomplete High-pressure Cylinder Expansion. 



-D 2 j V 

!. Diagram to Show Correct Low-pressure 
Cut-off for No Receiver Loss. 

high-pressure total volume DC, so 
that at release in the high-pressure 
cylinder the pressure would drop 
abruptly to such a value in the 
receiver as corresponds to filling 
the low pressure up to its cut-off, 
and work be lost equal to area 
CC'C". 

A shortening of the low-pressure 
cut-off will have an equally bad 
effect by introducing negative work 
as indicated in Fig. 74, in which 
the L.P. cut-off volume is reduced 
from DC to D'C'- Expansion in the 
high pressure proceeds as before till 
the end of the stroke, at which time 
it has a volume DC greater than 
the low pressure can receive D'C', 
hence the receiver pressure must 
rise to such a value as will reduce 
the volume the required amount, 
introducing the negative work CC'C" . 
Changes of low-pressure cut-off may, 
therefore, introduce negative work, 
change the receiver pressure and, 
of course, modifiy the distribution 
of work between high and low, but 



242 



ENGINEERING THERMODYNAMICS 



these same effects might also have resulted from changes of high-pressure 
cut-off or of cylinder ratio. 

For such conditions as have been assumed it seems that compounding 
does not increase the work capacity of fluids, but may make it easier to realize 
this capacity, introducing at the same time certain rather rigid relations between 
cut-offs and cylinder volumes as necessary conditions to its attainment. It 
can also be shown that the same proposition is true when there are clearance and 
compression, that is, in real cylinders and when the receiver is real or not 
infinite in size, or when the exhaust of high and admission of low, are not con- 
stant-pressure lines. The former 
needs no direct proof, as inspec- 
tion of previous diagrams makes 
it clear, but the latter requires 
some discussion. 

A real receiver of finite size 
is at times in communication 
with the high-pressure cylinder 
during its exhaust, and at other 
times with the low-pressure 
cylinder during admission, and 
these two events may take place 
at entirely independent times, be 
simultaneous as to time, or over- 
lap in all sorts of ways. Suppose 
the periods to be independent 
and there be no cylinder clear- 
ance, then at the beginning of 
high-pressure exhaust two sepa- 
rate volumes of fluid come 
together, the contents of both 
the high-pressure cylinder and the 
receiver, and this double volume 
is compressed by the H.P. piston 
into the receiver, in which case the 
high-pressure exhaust would take 
place with rising pressure . Follow- 
ing this will come low-pressure admission, during which the volume of fluid in the 
receiver expands into the low-pressure cylinder up to its cut-off, and if the same 
volume is thus taken out of the receiver as entered it previously, low-pressure 
admission will take place with falling pressure, the line representing it exactly 
coinciding with that for the high-pressure exhaust. Independence of H.P. 
cylinder exhaust and L.P. cylinder admission, as to time, may result in a cycle 
such as is represented in Fig. 75 for the case of no cylinder clearance. On this 
diagram the receiver line is DC, an expansion or compression line referred to a 
second axis of volumes KJ, placed away from the axis of purely cylinder volumes 




I i , .71. Diagram to Show Effect of Shortening L.P. 
Cut-off, Introducing a Receiver Loss Due to 
Over-expansion in the High-pressure Cylinder. 



WORK OF PISTON ENGINES 



243 



>y the distance LD, equal to the receiver volume to scale. All diagram points 
; are referred to the axis A I except those on the line DC. 

This same case of time independence of H.P. exhaust and L.P. admission 
yields quite a different diagram when the cylinder clearance is considered. 
Such a case is represented by the diagram, Fig. 76, which also serves to illustrate 
the effect of incomplete expansion and compression as to equalization of 
receiver with cylinder pressures. At high-pressue release the volume of fluid 
in the H.P. cylinder is ML and its pressure is LR. This is about to come 
into communication with the receiver volume ON from which the low-pres- 
sure cylinder finished taking fluid and which is, therefore, at the same pressure 



K', 



8r- 



. Receiver. 



V.olu 






FIG. 75. Diagram to Illustrate Variable Receiver Pressure for the Case of Independent 
High-pressure Exhaust and Low-pressure Admission with Zero Clearance. 

as the L.P. cylinder cut-off KS. The question, therefore, is what will be 
jthe pressure at P in both H.P. cylinder and receiver when LM cu.ft. of fluid 
ftt LR pressure combines with ON cu.ft. at pressure KS, and together occupy 
a volume ON+LM. By hypothesis the pressure after mixture is 

(first volume Xits pressure) + (second volume X its pressure) 
sum of volumes 



From this or the graphic construction following, the point P is located. If the 
high-pressure expansion had continued to bring LQ to the receiver pressure 
KS, it would reach it at X. At this hypothetic point there would be a volume 
NX in the H.P. cylinder to add to the volume ON in the receiver at the same 
pressure, resulting in OX cu.ft. This fluid would have a higher pressure at the 
[lesser volume of receiver and H.P. cylinder and the value is found by a compres- 



244 



ENGINEERING THERMODYNAMICS 



line through X, XPAT referred to the receiver axis. This same line is also 
the exhaust of the H.P. cylinder from P to A . A similar situation exists at admis- 
sion to the L.P. cylinder as to pressure equalization and location of admission 
line. At the end of the L.P. compression there is in the L.P. cylinder FE_ cu.ft. 
at pressure EH, to come into communication with the receiver volume CB cu.ft. 
at pressure BG, that at which H.P. exhaust ended. Producing the L.P. com- 
pression line to /, the volume BI is found, which, added to receiver, results 
in no pressure change. An expansion line, referred to the receiver axis through 7, 
fixes the equalized pressure J and locates the L.P. admission line JK, which, it 
must be observed, does not coincide with the H.P. exhaust. 

So far only complete independence of the time of H.P. exhaust and L.P. 
admission have been considered, and it is now desirable to consider the diagram- 




FIG. 76. Diagram to Illustrate Variable Receiver Pressure for the Case of Independent High- 
pressure Exhaust and Low-pressure Admission with Clearance. 

matic representation of the results of complete coincidence. Such cases occur 
in practice with the ordinary tandem compound stationary steam engine and 
twin-cylinder single-crosshead Vauclain compound steam locomotive. In the 
latter structure both pistons move together, a single valve controlling both 
cylinders, exhaust from high taking place directly into low, and for exactly 
equal coincident time periods. The diameter of the low-pressure cylinder being 
greater than the high, the steam at the moment of release suffers a drop in 
pressure in filling the low-pressure clearance, unless, as rarely happens, the pres- 
sure in the low is raised by compression to be just equal to that at H.P. release. 
After pressure equalization takes place, high-pressure exhaust and low-pres- 
sure admission events are really together a continuation of expansion, the 
volume occupied by the steam at any time being equal to the difference between 



WORK OF PISTON ENGINES 



245 



the two cylinder displacements and clearances up to that point of the stroke. 
Before this period of communication, that is, between high-pressure cut-off and 
release, the volume of the expanding fluid is that of the high-pressure displace- 
ment up to that point of the stroke, together with the high-pressure clearance. 
After the period of communication the volume of the expanding fluid is that 
of the low pressure cylinder up to that point of the stroke, together with the low 
pressure clearance, plus the high-pressure displacement not yet swept out, 
and the high-pressure clearance. 

These fluid processes cannot be clearly indicated by a single diagram, 
because a diagram drawn to indicate volumes of fluid will not show the volumes 
in the cylinders without distortion. If there be no clearance, Fig. 77 will 
assist in showing the way in which two forms of diagram for this purpose 
are derived. Referring to Fig. 77A, the volume A B admitted to the high 
pressure cylinder expands in it until it occupies the whole H.P. cylinder volume 
Z)C. At this point expansion proceeds in low and high together, with decreasing 
volumes in high and increasing in low until the low-pressure cylinder volume is 
attained at E. The line BCE then indicates the pressures and volumes of the 
fluid expanding, but does not clearly show the volume in either cylinder between 
C and E, with the corresponding pressure. It is certain, however, that when 
the volume in the H.P. cylinder becomes zero the pressure must have fallen 
to a value the same as that in the low when the fluid completely fills it, 
or P,=P.. 

As the high-pressure piston returns, on the exhaust stroke, the low- 
iressure piston advances an equal distance, on its admission stroke, sweep- 
ig through a greater volume than the high pressure, in the ratio of low-pres- 
re to high-pressure displacements. If at any point of the stroke the volume 

taining in the high-pressure cylinder be x, and the high- and low-pressure 
displacements be respectively DI, and I>2, then (Z>i x) is the volume swept 

out by high-pressure piston, x the volume remaining in it', and (Dix) 

the volume swept in by the low-pressure piston. Then the total volume 
still in the two cylinders is, for a point between C and E, 



ince the equation of the curve CE is, PV = P e Ve = PcDi, the value of V may 

Tie substituted, giving P \D 2 -x(j^l} =P c I>i,= constant. Dividing by 
I L \ i / J 

^-1 )this becomes PI T^-^ -x\=P\ ^^ z =a new constant, so 




246 ENGINEERING THERMODYNAMICS 

that if a new axis LM be laid off on B, 07= from the axis GP 



point on FC will be distant from the new axis LM an amount I (n 2 !_j^ ~ ) x \ 

as the product of this distance and the pressure P, is constant, the curve FC 
is an equilateral hyperbola referred to the axis LM. Therefore Fig. 77B is 




G 1 2 3 M 

COMBINED DIAGRAM TO ONE SCALE 

FIG. 77. Diagram to Illustrate the Compound Engine Cycle with No-receiver, and Exact 
Coincidence of H.P. Exhaust and L.P. Admission Periods. No Clearance. 

the pressure-volume representation of the entire cycle of the high-pressure 
cylinder. 

In Fig. 77C is shown the corresponding pressure-volume diagram for the low- 
pressure cylinder, for which it may similarly be shown that the curve DE may be 



WOEK OF PISTON ENGINES 24^ 

plotted to an axis NO at a distance to the left of the axis GP equal to the sam< 
quantity, 



AJ-ZV 



(295; 



These diagrams, 77 A, B and C may be superposed, as in Fig. 77 E, giving 
one form of combined diagram used for this purpose, and the one most nearly 
comparable with those already discussed. In this diagram, the area ABCFA 
represents the work of the high-pressure cylinder, and DEIHD, the work of 
the low-pressure cylinder. Together, they equal the work of the enclosing 
figure ABEIHA, and hence the work of the low-pressure cylinder must also be 
represented by the area FCEIHF. 

It is not difficult to show that if a vertical, CJ, be dropped from the point 
C to the exhaust line HJ, the figure CFHJ, in Fig. 77D is similar to DEIH, 
in Fig. 77E, reversed, but drawn to a different horizontal scale. Here the 
length of the low-pressure diagram is made equal to the length of the high- 
: pressure diagram. The two scales of volumes are shown above and below the 
figure. While this appears to be a very convenient diagram, it will be found 
to be less so when clearance and compression are considered. 

It may be noted that since it has been shown that the curves CF and DE 
are of the same mathematical form (hyperbolic) as the expansion line CE, 
they may be plotted in the same way after having in any way found the^axis. 
The location of this axis may be computed as given above, or may be found 
graphically by the method given in connections with the subject of clearance, 
Chapter I, and shown in Fig. 77J5. Knowing two points that lie on the curve, 
C and F, the rectangle CDFK is completed. Its diagonal, DK, extended, cuts 
the horizontal axis GV in the point M, which is the base of the desired axis ML. 
If now the axis NO and the figure DEIH, part of which is referred to this 
axis, be reversed about the axis GP, Fig. 77(7, NO will coincide with ML, 
Fig. 77D, and Fig. 78 results. Note that the axis here may be found graphically, 
in a very simple way. Draw the vertical CK from C to the axis GV and 
the horizontal DJ to the vertical IE extended. DC is then the high-pressure 
displacement and DJ the low-pressure displacement. Draw the two diagonals 
DK and JG, extending them to their intersection X. By similar triangles it 
may be shown that a horizontal line, UW, will have an intercept between these 
two lines, JG arid DK } equal to the volume of fluid present in the two cylinders 
combined. The intersection X is the point at which this volume would 
become zero if the mechanical process could be continued unchanged to that 
point, and is, therefore, on the desired axis ML extended. T being the inter- 
section of WU with the axis GP, when the volume UT is present in the high- 
pressure cylinder, TW gives the volume in the low-pressure cylinder. 

When clearance and compression are considered, this diagram is changed in 
many respects, and is shown in Fig. 79. The axes OP, 0V and 0V are laid out, 
with OZ equal to the clearance and ZK, the displacement, of the high-pressure 



ENGINEERING THERMODYNAMICS 



248 

cylinder, and OQ and QY, clearance and displacement of the low-pressure cylinder. 
It is necessary to know high-pressure cut-off, -=; high-pressure compression, 

; and low-pressure compression, =, hi addition to the initial and back 
ZK. *-( 

pressures, hi order to lay out the diagram. The drop in pressure CD at release 
is due to the coming together of (volume V c at pressure P c ), with (volume V, 
at pressure P 3 ). If the volume V,- (measured from axis OP) were decreased 
sufficiently to raise the pressure in the low-pressure clearance to the pressure 



























3 








































A 




B 








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pi 


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FIG. 78. Diagram to Show Volume of Steam in the Cylinders of the No-receiver Compound 
Engine at any Point of the Stroke for the Case of No-clearance and Coincidence of 
H.P. Exhaust with L.P. Admission. 

at C, the volume would become V s , as indicated at th,e point S } and the volumes 
now combined in the hypothetical condition would occupy the volume SC. 
Increasing this volume to D'D, that actually occupied after communication, 
the pressure would fall along the curve SD', which is constructed on KV and KT 
as axes. The pressure of D is then laid out equal to the pressure at D'. To 
find the axis, ML, for the curves DE and D'E', from any convenient point N 
on ZA, draw the line NK extended to X. Extend HG to R, at a height equal 
to that of N, and draw RQX, and through the intersection draw the desired axis 
XML. The fraction of stroke completed at E' in the low pressure at cut-off must 
be equal to that completed at E in the high pressure at compression, and may be 
laid out graphically by projecting up from E to the point U on the line NK and 
horizontally from U to W on the line RQ. Projecting down from W to the 
curve at E' locates the point of effective cut-off in the low-pressure cylinder. 



WORK OF PISTON ENGINES 



249 



kfter the supply from the high-pressure cylinder has been cut off at E', the 
xpansion is that of the volume in the low-pressure cylinder and its clearance, and 
ence the curve E'G is constructed on OP as an axis. 

While in this last case a zero receiver volume has been assumed, there is 
.othing to prevent a receiver volume being interposed between H.P. and L.P. 
o that common expansion takes place with a volume greater than assumed 
>y so much as this volume, the effect of which is to decrease the slope of DE 
nd D'E 1 '. Such receivers usually consist of the spaces included in a valve 
ody and connecting passages and may be treated generally as increased L.P. 
learances. 

The most common of all relations between H.P. exhaust and L.P. admis- 
ion is, of course, that of partial coincidence of periods, as it is thus with all cross- 




IG. 79. Diagram to Show Volume of Steam in the Cylinders of the No-receiver Compound 
Engine at any Point of the Stroke for the Case of Clearance and Coincidence of H.P. 
Exhaust with L.P. Admission. 

impound and triple-expansion engines having separate cranks for the 
^dividual cylinders. For these there is no simple fixed relation between the 
eriods, for, while crank angles are generally fixed in some comparatively simple 
Ration, such as 90, 180 and 270 for compounds and 120 for triples, they 
ie sometimes set at all sorts of odd angles for better balance or for better 
irning effort. Even if the angles were known the receiver line would have 
Ibe calculated point by point. When the H.P. cylinder begins to discharge into 
[receiver for, say, a cross compound with cranks at 90, steam is compressed 
ito the receiver, and so far the action is the same as already considered for 
dependence of periods, but at near mid-stroke the low-pressure admission 
fens while high-pressure exhaust continues. This will cause the receiver 
tessure to stop rising and probably to fall until the low pressure cuts off, which 



250 



ENGINEERING THERMODYNAMICS 



may occur before the H.P. exhaust into the receiver ceases. If it does, the receive: 
pressure will again rise. Exact determination of such complex receiver lines is 
not often wanted, and when needed is best obtained graphically point by point 
Its value lies principally in fixing exactly the work distribution between cylin- 
ders, which is not of great importance except for engines that are to work a1 
constant load nearly all the time, such as is the case with city water work* 
pumping, and marine engines. While equations could be derived for these 
cases, they are not worth the trouble of derivation, because they are too cumber- 
some, and graphic methods are to be substituted or an approximation to be made 
Four kinds of approximation are available, as follows, all of which ignon 
partial coincidence of periods: 

1. Receiver pressure constant at some mean value and clearance ignore 

2. Receiver pressure constant at some constant value and clearance co 
sidered with compression zero or complete. 

3. Receiver pressure fluctuates between fixed limits as determined by 
assumed size, clearance ignored. 

4. Receiver pressure fluctuates between fixed limits as determined by 
assumed size, clearance considered, with compression zero or complete. 

These are not all of equal difficulty in solution, and the one to be used 
that nearest the truth as to representation of conditions, which is usually t 
most difficult, provided time permits or the information is worth the troub 
Quickest work is accomplished with assumption (1) and as this is most oft 
used in practical work it indicates that its results are near enough for mo 
purposes. 

This discussion leads, therefore, to the analytical study of the following cycl< 

INFINITE RECEIVER, ZERO CYLINDER CLEARANCE. 
CYCLES V, AND VI (Fig. 80). 

(a) Admission at constant supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cyl. (may be zero) by law PV=c for (V 
PF*=cfor(VI). 

(c) Equalization of H.P. cyl. pressure with receiver pressu 

at constant volume (may be zero). 

(d) Exhaust into infinite receiver at constant pressure fro 

H.P. cylinder. 

(e) Equalization of H.P. cylinder pressure with supply pressu 

at constant zero volume. 

(f) Admission from receiver at constant receiver pressure 

L.P. cylinder. 

(g) Expansion in L.P cylinder (may be zero) by law PV = c f 

(V); PF'=cfor (VI). 
(h) Equalization of L.P. cylinder pressure with back pressu 

at constant volume (may be zero). 

(i) Exhaust at constant back pressure for L.P. cylinder. 
(f) Equalization of L.P. cylinder pressure with receiver pressi 

at constant zero volume. 



H.P. CYLINDER 
EVENTS 



L.P. CYLINDER 
EVENTS 



WORK OF PISTON ENGINES 



251 





























p 




































a I 


































a i 




















\ 


































| 


Cycles VII & VIII 
ycle VII PV=C Cycle VIII PV=C 




\ 


Cycles V & VI 
Cycle V PV=C Cycle VI PVS=C 














e 






\ 
















\ 


















t 


I 


V 






























\ 




\ 




















V 
































SB 


V 




















o 


\ 
















-< 


Rec 


ivt'i 


Vol 


line 


-* 




l\ 


c 

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x 


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k 














































































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V 




















































































































a I 


















p 
















a \ 




















B 


Cycles \X&X 
Cycle IX PV=C Cycle X 


PV=C 














f 


1 


Cycles XIII & XIV 






\ 
















\Cycle XIII PV=C Cycle XIV PV=C 


d 




5 


































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\ 




































\ 
















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,/ 


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Rec 


eivei 


Vol 


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x v 




























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a 11 






















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Cycles XI & XII 
Cycle XI PV=C Cycle XII PV=C 













Cycles XV & XVI 
Cycle XV PV=C Cycle XVI PVS=C 










l\ 






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A 




































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FIG. 



V V 

go. Compound Engine Standard Reference Cycles or PV Diagrams. 



252 



ENGINEERING THERMODYNAMICS 



RELATIONS 

BETWEEN H.P. 

AND L.P. CYLINDER 

EVENTS 



(1) H.P. exhaust and L.P. admission independent as to time, 

coincident as to representation (except as to length). 

(2) H.P. expansion line produced coincides as to representation 

with L.P. expansion line. 

(3) The length of the constant pressure receiver line up to the 

H.P. expansion line produced is equal to the length of 
the L.P, admission line. 



H.P. CYLINDER 
EVENTS 



FINITE RECEIVER, ZERO CYLINDER CLEARANCE. 
CYCLES VII, AND VIII, (Fig. 80). 



(a) Admission at constant supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cylinder (may be zero) by law PV=c 
for (VII); PV s =c for (VIII). 

(c) Equalization of H.P. cylinder pressure with receiver pressure 

at constant volume (may be zero) with a change of receiver 
pressure toward that at H.P. cylinder release (may be 
zero). 

(d) Exhaust into finite receiver from H.P. cylinder at rising 

pressure equivalent to compression of fluid in H.P. cylinder 
and receiver into receiver by law PF=c for (VII) and 
PV s =c for (VIII). 

(e) Equalization of H.P. cylinder pressure with supply pressure 

at constant zero volume. 



L.P. CYLINDER 
EVENTS 



RELATION 

BETWEEN H.P. AND 

L.P. CYLINDER 

EVENTS 



(/) Admission from receiver to L.P. cylinder at falling pressure 
equivalent to expansion of fluid in receiver into receiver 
and L.P. cylinder together by law PV=c 'for (VII), PV S =c 
for (VIII). 

(0) Expansion in L.P. cylinder (may be zero) by law PV=c 

for (VII); PV s =c for (VIII). 

(h) Equalization of L.P. cylinder pressure with back pressure 
at constant volume (may be zero). 

(1) Exhaust at constant back pressure for L.P. cylinder. 

(/) Equalization of L.P. cylinder pressure with receiver pressure & 
at constant zero volume to value resulting from H.P.(y 
exhaust. 

(1) H.P. exhaust and L.P. admission independent as to time,! 

coincident as to representation, except as to length. 

(2) H.P. expansion line produced coincides as to representation!! 

with L.P. expansion line. 

(3) The length of the receiver pressure line up to the H.P.I 

expansion line produced is equal to the length of the | 
L.P. admission line. 



WOEK OF PISTON ENGINES 



253 



No RECEIVER, ZERO CYLINDER CLEARANCE. 
CYCLES IX, AND X, (Fig. 80). 



H.P. CYLINDER 
EVENTS 



BOTH H.P. AND L.P. 
SIMULTANEOUSLY 

H.P. CYLINDER 

EVENTS 



L.P. CYLINDER 
EVENTS 



(a) Admission at constant supply pressure to H.P. Cylinder. 
(6) Expansion in H.P. cylinder (may be zero) by law PV 
for (IX); PF s =cfor(X). 



(c) Transference of fluid from H.P.to L.P. cylinder with simul- 
taneous continuation of expansion until all fluid is in 
L.P. cylinder and expanded to its full volume by law 
PV=c for (IX); PV S =c for (X). 

( (d) Equalization of H.P. cylinder pressure to the pressure of 
supply. 

(e) Equalization of L.P. cylinder pressure with back pressure 

at constant volume (may be zero). 
(/) Exhaust at constant back pressure for L.P. cylinder. 
(g) Equalization of L.P. cylinder pressure to the pressure in 

H.P. cylinder at the end of its expansion. 



[.P. CYLINDER 
EVENTS 



INFINITE RECEIVER, WITH CYLINDER CLEARANCE. 
CYCLES XI, AND XII, (Fig. 80). 

(a) Admission at constant supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cylinder (may be zero) by law PV=c 
for (XI); PF s =cfor(XII). 

(c) Equalization of H.P. cylinder pressure with receiver pressure 

at constant volume (may be zero) pressure. 

(d) Exhaust into infinite receiver at constant pressure from H.P. 

cylinder. 

(e) Compression in H.P. cylinder to clearance volume (may be 

zero) by law PV =c for (XI) ; PF S =c for (XII). 
(/) Equalization of H.P. cylinder pressure with supply pressure 
at constant clearance volume, may be zero. 

(g) Admission from receiver at constant-receiver pressure to 
L.P. cylinder. 

(ft) Expansion in L.P. cylinder (may be zero) by law PV c 
for (XI); P7 s =cfor(XII). 

(i) Equalization of L.P. cylinder pressure with back pressure 
.P. CYLINDER I at constant volume (may be zero). 

EVENTS | (/) Exhaust at constant back pressure for L.P. cylinder. 

(k) Compression in L.P. cylinder to clearance volume by law 
PV = c for (XI) ; PV S =c for (XII) (may be zero). 

(/) Equalization of L.P. cylinder pressure with receiver pres- 
sure at constant clearance volume without change of 
receiver pressure (may be zero). 



254 



ENGINEERING THERMODYNAMICS 



RELATIONS 

BETWEEN H.P. AND 

L.P. CYLINDER 

EVENTS 



(1) H.P. exhaust and L.P. admission independent as to til 

coincident as to representation except as to length. 

(2) L.P. expansion line does not coincide as to representation 

with H.P. expansion line produced by reason of clearance 
influence except in one special and unusual case. 

(3) The length of the constant-receiver pressure line intercepted 

between H.P. compression line and H.P. expansion line 
produced is equal to the same intercept between L.P. 
expansion line and L.P. compression line produced. This 
is equivalent to the condition that the volume taken in by 
low is equal to expelled by the high reduced to the same 
pressure. 



FINITE RECEIVER, WITH CYLINDER CLEARANCE. 
CYCLES XIII, AND XIV, (Fig. 80). 



H.P. CYLINDER 
EVENTS 



L.P. CYLINDER 
EVENTS 



(a) Admission at constant supply pressure to H.P. cylinder. 
(6) Expansion in H.P. cylinder (may be zero) by law PV=c 
for (XIII) ; PV S =c for (XIV). 

(c) Equalization of H.P. cylinder pressure with receiver pres- 

sure at constant volume (may be zero) toward that at 
H.P. cylinder release (may be zero). 

(d) Exhaust into finite receiver from H.P. cylinder at rising 

pressure equivalent to compression of fluid in H.P. cylinder 
and receiver into receiver by law PV =c for (XIII) ; PV S =c 
for (XIV). 

(e) Compression in H.P. cylinder to clearance volume (may be 

zero) by law PV=c for (XIII); PV S =c for (XIV). 
(/) Equalization of H.P. cylinder pressure with supply pressure 
at constant clearance volume. 

(g) Admission from receiver to L.P. cylinder at falling pressure; 

equivalent to expansion of fluid in receiver into receiver j 

and L.P. cylinder together by P7=c for (XIII); PV S =c\ 

for (XIV). 
(h) Expansion in L.P. cylinder (may be zero) by law PV =c\ 

for(V); PV =cfor(VI). 
(i) Equalization of L.P. cylinder pressure with back pressure! 

at constant volume (may be zero). 

(f) Exhaust at constant back pressure for L.P. cylinder. 

(k) Compression in L.P. cylinder to clearance volume by lawj 

PV=c for (XI); PV'=c for (XII) (may be zero). 
(/) Equalization of L.P. cylinder pressure with receiver pressure j 

at constant clearance volume with change of receive! |; 

pressure in direction of L.P. compression pressure (maj) 

be zero). 



WORK OF PISTON ENGINES 



255 



RELATIONS 

ETWEEN H.P. AND 

L.P. CYLINDER 

EVENTS 



(1) H.P. exhaust and L.P. admission independent as to time, 

representation and length. 

(2) L.P. expansion line does not coincide as to representation 

with H.P. expansion line produced by reason of clearance 
influence except in one special and unusual case. 

(3) The high-pressure exhaust and low-pressure admission lines 

do not coincide as to representation by reason of clearance 
influences. 

(4) There is a relation between the lengths of the L.P. admission 

and H.P, exhaust lines, but not a simple one. 



No RECEIVER, WITH CYLINDER CLEARANCE. 
CYCLES XV, AND XVI, (Fig. 80). 



H.P. 

CYLINDER 

EVENTS 




LT. 

CYLINDER 

EVENTS 



(a) Admission at constant-supply pressure to H.P. cylinder. 

(6) Expansion in H.P. cylinder (may be zero) according to law 

PV=c for (XV); PV S =c for (XVI). 
(c) Equalization of pressures in H.P. cylinder after expansion 

with that in L.P. after compression at constant volume 

(may be zero) . 
Transference of fluid from H.P. to L.P. cylinder until all 

fluid is in L.P. cylinder and expanded to its full volume 

by same law as (6) . 
Compression in H.P. cylinder to clearance volume (may be 

zero) by law PV=c for (XV); PV S =c for (XVI). 
(/) Equalization of pressure in H.P. cylinder with supply at 

constant-clearance volume (may be zero). 

(g) Expansion in L.P. cylinder may be zero by law PV=c for 

(XV); FF s =cfor(XVI). 
(h) Equalization of pressure in L.P. cylinder with back pressure, 

at constant volume (may be zero). 
(i) Exhaust at constant pressure for L.P. cylinder, 
(j) Compression in L.P. cylinder to clearance, may be zero by 

law PV =c for (XV) ; PV 9 =c for (XVI). 
(k) Equalization of L.P. cylinder pressure with H.P. cylinder 

pressure. 



Cycle XVII, Fig. 81, for the triple expansion is denned in the same way 
.s the corresponding case of compounds Cycle V, with appropriate alterations 
9 wording to account for a third or intermediate cylinder between high- and low- 
pessure cylinders and an additional receiver. Thus, high- pressure cylinder- 
xhausts into first, and intermediate cylinder into second receiver: inter- 
mediate cylinder receives its supply from first, and low-pressure cylinder from 
|cond receiver. This being the case, it is unnecessary to write out the cylin- 
mr events, noting their relation to the corresponding compound case. 



256 



ENGINEERING THERMODYNAMICS 



9. Compound Engine with Infinite Receiver, Logarithmic Law. No Clear- 
ance, Cycle V. General Relations between Pressures, Dimensions, and Work. 
It must be understood that the diagrams representing this cycle, Fig. 82, 
indicating 'A) incomplete expansion and (B) over-expansion in both cylinders, 

P 



m 



FIG. 81. Triple-expansion Engine Standard Reference Diagram or PV Cycle for Infinit 

Receiver. 



may just as well stand for over, complete or incomplete expansion in all possibl 
combinations in the two cylinders. Applying the principles already derivec 
for calculating the work areas, 

High-pressure cylinder work 



fefc-w* 



(296 



Low-pressure cylinder work 



(297 



WORK OF PISTON ENGINES 257 

Total work 

P d V d -P V ff> . . (298) 



pressure being in pounds per square foot, and volumes in cubic feet. 

In theses expressions the receiver pressure P e = P d is unknown, but determinate 
as it is a function of initial pressure and certain volumes, giving it the value, 



merely satisfying the condition that the point E at which expansion begins 
in the low-pressure cylinder must lie in the expansion line of the high. Sub- 
stituting this value there results 



?+lQ fe --PJV . (299) 



This is a perfectly general expression for the work of the fluid expanding to 
any degree in two cylinders in succession when the clearance is zero and receiver 
volume infinite, in terms of initial and back pressures, pounds per square foot, 
the volumes occupied by the fluid in both cylinders at cut-off, and at full 
stroke in cubic feet. Dividing this by the volume of the low-pressure cylinder 
V gives the mean effective pressure referred to the low-pressure cylinder, 
from which the horse-power may be determined without considering the high- 
1 pressure cylinder at all. Hence, in the same units as are used for P b and P g , 



F*r F F/ F/l 

(M.E.P. referred to L.P.) =Pt,~\ 2+loge ^+loge -^^ -P 9 . 

Vol Fft Ye Ye] 



(300) 



Proceeding as was done for simple engines, the work per cubic feet of fluid 
supplied is found by dividing Eq. (299) by the volume admitted to the high- 
ipressure cylinder F/,, whence, 



Workpercu.ft. supplied = p2+log e +log e --P. . . (301) 



Also applying the consumption law with respect to horse-power, 

13 750 F 

:Cu.ft. supplied per hour per I.H.P. = ^ to , ow) ^ ..... (302) 

Lbs. supplied per hr. per I.H.P. = (m g p * Ki fo lqw) pfri ..... (303) 



258 



ENGINEERING THERMODYNAMICS 



These last five equations, (299), (300), (301), (302), (303), while character- 
istic, are not convenient for general use in their present form, but are ren- 




( rel.pr. ) L 

( atn>.pr.> 
Q(bk.pr.) 



H.P. 




L.P. 



IN DICATOR CARDS OF 



H.P 



? Cbk.pr.) 

7 (rel.pr.) L 

(atni.pr.) 



L.P. 



INDICATOR CARDS OF 

EQUAL BASE AND 

HEIGHT 



H V 



FIG. 82. Work of Expansive Fluid in Compound Engine with Infinite Receiver, Zero 

Clearance. Cycle V, Logarithmic Expansion and Cycle VI Exponential Expansion. 

dered so by substituting general symbols for initial and back pressures 

displacement, cut-off, and amount of expansion for each cylinder. 



WORK OF PISTON ENGINES 



259 



.",et (in.pr.) = initial or supply pressure, pounds per square inch = T-A; 

144 

1 ' (rel.pr.)# = release pressure, in H.P. cylinder pounds per square inch = 



p 



p 

(rel.pr.)z,= release pressure in L.P. cylinder, pounds per square inch= ~ ; 

(rec.pr.) = receiver pressure, pounds per square inch=:^ = :rA; 

p 

(bk.pr.) = back pressure, pounds per square inch=Y~; 

RH = ratio of expansion in high-pressure cylinder = ^\ 

' b 

R L = ratio of expansion in low-pressure cylinder \ 



R v = ratio of expansion for whole expansion = 



Vt> 



D H = displacement of high-pressure cylinder = V d = V c ; 
1 D L = displacement of low-pressure cylinder =V f =V g ; 

1 R c = cylinder ratio = -=f = -=/ ; 

DH Vd 

' Z H = fraction of displacement completed up to cut-off in high-pressure 

' cylinder, so that Z H D H =V b = ^; 

tlH 

i Z L = fraction of displacement completed up to cut-off in low-pressure 

cylinder, so that Z L D L =Ve=-p-' 

KL 

ibstitution of these general symbols in Eqs. (299), (300), (301), (302), and 
(303) gives another set of five equations in useful form for direct substitu- 
tion of ordinary data as follows : 

Work of cycle 



. (304) 



W -]^r - 144(bk.pr.)l>L (a) 

= 144A, I (in.pr.)^(2+log, y- +lo& |- ~^V) ~ ( bk -P r - 
Jtic\ AH AL, HC&L/ 

(in.pr.)^-^l 2+loge RH+IO& X L -~] - (bk.pr.) j. (c) 



(m.e.p.) Ibs. per sq.in. referred to L.P. cyl. 
= (in.pr . 



= (in.pr .)~(2+loge ^-+log e -sr - nrp-J ~ (bk.pr.) (b) 
'#A ^// *L ZLKC/ J 



. (305) 



260 



ENGINEERING THERMODYNAMICS 



Work per cu.ft. supplied 

= 144 T (in.pr.) (2+loge/i 

[/ 1 1 

(in.pr.) ( 2+logc =- +loge -=r- 
\ H L 

Cu.ft. supplied per hr. per I.H.P. 

13,750 



(a) 



(306) 



(m.e.p. ref. to L.P.) R H Rc 
13,750 Z H 



(a) 



(307) 



J 



(m.e.p. ref. to L.P.) R c 

From this, of course, the weight in pounds supplied per I.H.P. results directly 
from multiplication by the density of the fluid. 

To these characteristic equations for evaluating work, mean pressure, 
economy and consumption in terms of the initial and final pressures and cylin- 
der dimensions there may be added a series defining certain other general rela- 
tions of value in fixing the cycle for given dimensions and initial and final 
pressures, and in predicting dimensions for specified total work to be done and 
its division between high- and low-pressure cylinders. 

Returning to the use of diagram points and translating into the general 
symbols as each expression is derived, there results, 



Receiver pressure = P d = P e = Pb- 






(in.pr.)^% (6) 



. . . (308) 



High-pressure cylinder release pressure = P c = , 

.'. (rel.pr.)* = (in.pr.) 
KM 

= (in.pr.)Z* 



y 
Low-pressure cylinder release pressure = P f =P e ~. 

i 
.'. (rel.pr.) L = (in.pr.) 



R v 

(rol.pr.)w 
Rc 



(a) 
(6) 

(a) 
(6) 
(c) 
(d) 



. (309) 



(310) 



WORK OF PISTON ENGINES 



261 



Division of work between cylinders may be made anything for a given load by 
suitably proportioning cylinders, and equations giving the necessary relations 
to be fulfilled can be set down. It is quite common for designers to fix on 
equal division of work for the most commonly recurring or average load or that 
corresponding to some high pressure cut-off or low-pressure terminal pressure, 
generally the latter. Therefore, a general expression for dimensional rela- 
tions to be fulfilled for equal division of work is useful. On the other hand, 
for an engine the dimensions of which are determined, it is often necessary to 
find the work division for the imposed conditions, so that the following equa- 
tions are of value. 

From Eqs. (296) and (297), noting that P d = P e = P &1 /, 



High-pressure cylinder work _ 
Low-pressure cylinder work ~ 



PtF 



1+log> ^ _ 



1+loge RH~^ 

tic 



(a) 



1 1 

1+loge -~- 



. . (311) 



This is a general expression for work division between the cylinders in 
terms of (a) ratio of expansion in each cylinder, initial and back-pressure 
ratio and cylinder ratio, or, in terms of (6) cut-off in each, associated with cylinder 
and pressure ratios. 

This expression Eq. (311) is less frequently used in its general form as above, 
than in special forms in which the work of the two cylinders in made equal, 
or the expression made equal to unity. The conditions thus found for equal 
division of work between cylinders may be expressed either (a) in terms of initial 
and back pressures, release pressure of low-pressure cylinder and ratio of L.P. 
admission volume to H.P. displacement, and cylinder ratio, or (6) cut-off in 
high- and low-pressure cylinders, initial and back pressures and cylinder ratio. 
Still more special conditions giving equality of work may be found (c) when the 



262 ENGINEERING THERMODYNAMICS 

cylinder ratio is made such that equality of work is obtained at all loads by 
equalizing high and low cut-offs. 

(a) To find the first set of conditions, equate Eqs. (296) and (297) from the 
first part of this section, and by simplification there results, 



or 

~ ~ ^V e 



Introducing the usual symbols and putting in addition 

Low-pressure admission volume _ Ve _ _ _ (rel.pr.)// 
High-pressure displacement volume V c (rec.pr.) 

Therefore, 

&?L>^_11 

(rel.pr.W x J 



[ L(rel.pr.) L xj (in.pr.) 1* 

Rc= \e ; , N x\ . 

(rel.pr.) L J 






This is of value when a given release pressure is to be reached in the low 
pressure cylinder and with a particular value of low-pressure cut-off volume 
as fixed by x in terms of high-pressure cylinder displacement. 

(b) Again for equal division of work, make Eq. (311) equal to unity, 
whence, 



1+Io& _ 

ZIH KC&L 



or 



which may be reduced to the following, solving for R c , 

^ = ~^j^^T ~' ' ' (318) 

Z H (in.pr.) 

Equal division of work for given initial and back pressure is to be obtained 
by satisfying these complex relations Eq. (313) between the two cut-offs, or their 
equivalent ratios of expansion in connection with a given cylinder ratio, or the 
relation between pressures and volumes in Eq. (312) equally complex. 



WORK OF PISTON ENGINES 



263 



(c) An assumption of equal cut-off in both cylinders gives results which are 
of .interest and practical value, although it is a special case. Eq. (313) then 
becomes, when ZH ZL or R H = RL, 



As would be expected, this may also be derived from Eq. (312), since 

1 bk.pr. (rel.pr.)z, 

-=, , v and x= ,, . ^ \ under these conditions. 

x (rel.pr.)z, (bk.pr.) 

The receiver'pressure under these conditions is constant, and is, from Eq. (308) 



(in.pr.) (in.pr.) 

/ r> / \ 

KG /in.pr. y 
\bk.pr./ 



.(315) 



The high-pressure release pressure is not affected by any change in the low- 
pressure cut-off, and hence Eq. (309) gives the value of high-pressure release, 
pressure for the case. Low-pressure release pressure Eq. (310) may be expressed 
for the case of equal cut-off, 



/ r \*7 V ~\ i ' 

(tel.pr.)i = y^^^ = ZJ (in.pr.) (bk.pr.) (a) . . 



= ^- (in.pr.) (bk.pr.) 
tin\~ 



(6) . . . . 



(316-) 



The foregoing equations up to and including Eq. (311), are perfectly general, 
and take special forms for special conditions, the most important of which is 
that of complete expansion in both cylinders, the equations of condition for which 
are, referring to Fig. 82. 



Pf=p , 

which, when fulfilled, yield the diagram, Fig. 83. These equations of condi- 
tion are equivalent to fixing a relation between the cut-off in both the high- 
and low-pressure cylinders, and the volume of high-pressure cylinder with respect 
to the low-pressure volume, so that 



Pf 

V b = Vf ~, or symbolically, 

*b 

^/bJc.prA /bk.prA ' 
Da \rn.pr. / \in.pr. / 

R " = R~ c \tiS^) {b 



(317) 



264 



ENGINEERING THERMODYNAMICS 



Similarly the low-pressure cylinder cut-off volume must equal the high- 
pressure displacement volume or D H = 



DM 1 



(6) 



(318) 



indicating that low-pressure cut-off is the reciprocal of the cylinder ratio. Making 
the necessary substitution there result the following equations for this cycle 
which, it must be noted, is that for most economical use of fluid in compound 



* z 



Z L D r 



(i 



IT.) 



( r c.pr.) 




N C 



E(rel,pr.) 



FIG. 83. Special Case of Cycles V and VI, Complete Expansion in both Cylinders of 
Compound Engine with Infinite Receiver and Zero Clearance. 

cylinders without clearance and with infinite receiver, and in which the same 
work is done as in Cycle I, for simple engines at best cut-off. 
From Eq. (308) 



. . . (319) 



From Eq. (309), 






WOEK OF PISTON ENGINES 



265 



From Eq. (310), 



x (in.pr.) (in.pr.) ,, . x 

.pr.)z, = Vp = i / ' x = (bk.pr.) 
KcKH yin.pr.; 



(321) 



p _ 
Kc 



Rc (bk.pr.) 



From Eq. (311), 



High 


pressure 


cylinder 


work 


Re 


loge RH 


(a) 

/1A 


Low 


pressure 


cylinder 


work . 


og^-ga^w 


loge RL 

1 



. (322) 



For the case of most economical operation, that of complete and perfect 
expansion in both cylinders, there may be set down the four characteristic 
Eqs. (304), (305), (306), (307) with suitable modifications to meet the case. 
These become 



Work of cycle = 144(in.pr.)Dzr 



.pr 



/in.pr. \ 
Vbk.pr./ 



bk.pr 



^Z. . . (323) 



. ,(324) 



R 



* 

1 Work per cu.ft. supplied = 144(in.pr.) log e ~ = 144(in.pr.) log e R v . (325) 



. . . . (226) 



13,750 /bk.prA 

Cu.ft. per hr. per I.H.P. = 7 - , . T P X(T -) (a 

(m.e.p. ref. to L.P.) \in.pr. / 



13,750 



(m.e.p.ref . to L.P.) Rv 



(b) 



ijor .equal division of work with complete expansion in both cylinders, the 
ratios of Eqs. (317) and (318) becomes 



bk.pr A * 






(327) 



id this is evidently a case to which Eqs. (314) and (315) apply without change. 



266 ENGINEERING THERMODYNAMICS 

Example. 1. Method of calculating diagram, Fig, 82. 
Assumed data for Case A: 

p a = p b = 100 Ibs. per sq.in. abs. V a = V n = V m = cu.ft. 

p n= p d= p e= 50 Ibs. per sq.in. abs. V c = V d = .6 cu.ft. 

p m =p g = 10 Ibs. per sq.in. abs. V f = V g = 2 cu.ft. 

P c = 60 t lbs. per sq.in. abs. V e = .8 cu.ft. 

To obtain point B: 

p ar\ 

Vb = VcX ^. = ,Q X =.36cu.ft. 
* b 

To obtain point F: 

=201bs. persq.in. 



Vf 

To construct the indicator cards: 

Lay off ND of the PV diagrams to equal the length of the card, and NA perpen 
dicular to it at N to equal the height of the card. Cut off equals AB+ND. From 
on card lay off this ratio times the length of the card. From D on the card lay off 
a perpendicular equal to CD of the PV diagram reduced by the same proportioi 
as AN of the card is to AN of the diagram. Join the points B and C by a curv 
through points located from intermediate points on the PV diagram. The low 
pressure card is constructed in same manner. 

Example. 2. A 12- and 18x24-in. steam engine without clearance runs on 15i 
Ibs. per square inch absolute initial pressure, 10 Ibs. per square inch absolute back pressure 
and has a speed of 125 R.P.M. What will be (a) the horse-power for cut-off in H.P 
cylinder, (b) pounds of steam per I. H.P. hour, (c) terminal pressures, (d) L.P. cut-off 
for continuous expansion, (e) work done in each cylinder. 

NOTE: B for 150 Ibs. =.332. 

(a) From Eq. (305) 
(m.e.p.) referred to L.P, cylinder is 



(in.pr.) I 2+log e J?//+log e #z, ] (bk.pr.). 

KnKc \ KC/ 

In this case 

I) =2.25, ^=2.25, 



since vol. of L.P. cyl. at cut-off must be equal to the entire volume of the high fo 
continuous expansion, hence 

(m.e.p.)=150XX(2+.69+.81-l)-10=73.3 Ib. sq. inch, 



22 
and 



(b) From Eq. (307) 



WORK OF PISTON ENGINES 267 

(c) From Eq. (309) 



(rel.pr.)# = (in.pr.)^a-, 

= 150X^=75 Ibs. sq.in, 



and from Eq. (310) we have 



75 
= =33.3 Ibs. sq.in. 



(e) From Eq. (311) 

1 1 

H.P. work ZH RcZL 



L.P. work (bk.pr.) R c ' 



1+.69- 



2.25 X. 44 .69 , Bfl 
- = .456, 



81- - L51 

H.P.work =. 456 XL.P. work, 

f| | H.P. work+L.P. work =282 I.H.P. 

ce 

H.P. work =88 I.H.P. 

L.P. work = 194 I.H.P. 

Prob. 1. What must be the cylinder diameters of a cross compound engine to run 

100 Ibs. per square inch absolute steam pressure, 18 ins. of mercury vacuum and to 
Develop 150 H.P. at a speed of 200 R.P.M. with \ cut-off in each cylinder, if cylinder 
mtio is 3 and stroke is 18 ins.? Engine is double-acting and assumed to have no 
clearance. 

Prob. 2. What will be the release pressure in each cylinder and the receiver 
aressure of the engine of Prob. 1? If cut-off were reduced to | in H.P. cylinder, 
few would these pressures be affected and to what extent? How would the horse- 
power change? 

Prob. 3. A 15- and 22 X30-in. infinite receiver engine has no clearance, a speed of 
J50 R.P.M., initial pressure 125 Ibs. per square inch gage. What will be the horse- 
iower and steam consumption for a H.P. cut-off of J, i, f , \, and that value which 
fill give complete expansion in high-pressure cylinder? Low-pressure cut-off to be 
ixed at \. 
| NOTE: $ for 150 Ibs. gage = .363. 



268 ENGINEERING THERMODYNAMICS 

Prob. 4. What will be the release and receiver pressures, and the work done in each 
cylinder for Prob. 3? 

Prob. 5. An 18 and 24x30-in. infinite receiver engine is to be operated so as 
to give complete expansion in both cylinders. What will be the cut-off to accomplish 
this and what horse-power will result if the initial pressure is 100 Ibs. and back pressure 
10 Ibs. per square inch absolute? 

Prob. 6. Draw the PV diagram for following cases. Cylinder ratio 1 to 2.5, 
(in.pr.), 100 Ibs. per square inch absolute, (bk.pr.), 20 Ibs. per square inch absolute, 
H.P. cut-off (a)=l (6)=J, (c)=i L.P. cut-off (a) =4, (6) =-&, (c) =1 . 

Prob. 7. For the following conditions find the horse-power, steam used per hour, 
receiver pressure and release pressures. Engine, 10- and 15x24-in. 150 R.P.M., 
125 Ibs. per square inch gage initial pressure, 2 Ibs. per square inch absolute, back 
pressure, \ cut-off in high-pressure cylinder, M> cut-in low-pressure cylinder with 
infinite receiver. 

NOTE: 8 for 125 Ibs. =.311. 

Prob. 8. An infinite receiver engine is to develop 150 H.P. at 200 R.P.M. when 
initial pressure is 150 Ibs. per square inch absolute. Cylinder ratio is 1 to 3 and 
back pressure is one atmosphere. What must be its size if the stroke is equal to 
the low-pressure cylinder diameter for \ cut-off in the high-pressure cylinder and 
| cut-off in the low-pressure cylinder? 

Prob. 9. Find by trial the cut-offs at which work division will be equal for an 
infinite receiver engine with a cylinder ratio of 2.5, an initial pressure of 100 Ibs. pei 
square inch absolute and a back pressure of 5 Ibs. per square inch absolute? 

10. Compound Engine with Infinite Receiver, Exponential Law. Nc 
Clearance, Cycle VI. General Relations between Pressures, Dimensions 
and Work. Again referring to Fig. 82, which may be used to represent thi: 
cycle also, the work of each cylinder may be expressed as follows, by th 
assistance of Eq. (254) derived in Section 4. 



. . . (3281 
. . . (3291 



where Z H is the cut-off in the high pressure, = ~ and Z L , low-pressure cut-o: 

V e 
= Y' In combining these into a single equation for the total work, the terr 

for receiver pressure (rec.pr.) should be eliminated. Referring to Fig. 82, 

(*.)-P.-P.-P$)'-<tor.)(-)', . . . 
hence 



WORK OF PISTON ENGINES 269 

a rather complex expression which permits of little simplification, but offers 
no particular difficulty in solution. 

Mean effective pressure referred to the low-pressure cylinder is 



-<*-> 



Work per cubic feet fluid supplied may be found by dividing Eq. (331) by 
the supply volume, which in terms of low-pressure displacement is 

(Sup.Vol.)=D^ (333) 

The consumption of fluid, cubic feet per hour per indicated horse-power is 
jConsumption cu.ft. per hr. per I.H.P. = -; ' ^, . . . (334) 

which is the same expression as for the logarithmic law. Multiplying this by 
81, the initial density of the fluid, pounds per cubic feet, gives consumption, 

^pounds per fluid hour per I.H.P. 

S| ! The receiver pressure has already been determined in Eq. (330). 

[isj Release pressure of the high-pressure cylinder is 

(rel.pr.)*=(in.pr.)Z* s , . . V ; V ! . ; . . (335) 
for the low-pressure cylinder, 

(rel.pr.)z,= (in.pr.)(|?Y (a) 



(336) 
(m.pr.) 



rhere R v is the ratio of maximum volume in the low pressure, to, volume at 

r> 

jut-off in the high, and equals =. 

AH 

The distrubition of work between the high- and low-pressure cylinders may 
>e found as follows, by means of Eqs. (328) and (329), eliminating (rec.pr.) by 
{leans of Eq. (330) 



s-l / \RcZj 

TT7 r / r7 \ / ~ r/ a \ \ /U1r- * \ 1 " * * v ' / 

W L r 



270 



ENGINEERING THERMODYNAMICS 



Equality of work in the two cylinders will be obtained if this expression 
is equal to unity, giving a complex relation between high- and low-pressure 
cut-offs, cylinder ratio and ratio of initial and back pressures, to be satisfied. 
It is found at once that the simple conditions for equality in the case of logarith- 
mic law will not give equality of work for the exponential law. There is, how- 
ever, a case under this law which yields itself to analysis, that of complete expan- 
sion in both cylinders, without over-expansion. The conditions for equality 
of work for this case will be treated after deriving work and mean effective 
pressure for it. 

Complete expansion, without over-expansion, in both cylinders may be 
represented by Fig. 83. 



AB 



NC 



and since 



P =___ 
DH NC Z L . 



The true ratio of expansion = J R F =^| = 1 =|^ J but this is also equal to 

AB AH&L AH 

\F ^r") * dUe t0 the laW f th 



By means of Eq. (257) in Section (4) the work of the two cylinders may be 
evaluated, 




(a) 



(338), 



but since 



(339 



I 



WORK OF PISTON ENGINES 



271 



The total work is evidently the same as that of a cylinder equal in size to 

i? 
the low-pressure cylinder with a cut-off equal to ~, working between the 

He 

given (in.pr.) and (bk.pr.) and may be stated by reference to Eq. (257), 
Section 4, or by taking the sum of W H and W L given above, 



which reduces to 



Z H s 

'Rcs-l 



{ fc 



(340) 



For this case of complete expansion in both cylinders, the ratio of high- to 
How-pressure work is given by division and cancellation, 



TF 

W 



L ( V 

\Rc) 



f- -Re" 

AH/ 

Rc^-l ' 



(341) 



Equality of work, obtained by placing this expression equal to unity, pro- 
ivides the condition that 



s-l 



*- 



Re 



equal work and complete expansion, and 



s-l 



Z H = 



bk. 



.pr s \n.pr. 

= ~ 



1 

:-l 



(342) 



(343) 



Since Z L = for complete expansion, 'and (in Fig. 83) P c V c s = PfV f s , the 
iiver pressure, P c , is 

(rec.pr.) = (bk.pr.)(^y=(bk.pr.)fl<7', !? . , ',. . . (344) 

which Re will have the value given above if work is equally distributed. 

Example 1. What will be (a) the horse-power, (6) consumption, (c) work ratio, 
receiver and release pressures for the following conditions? Engine 12 and 

X24 ins., running at 125 R.P.M. on initial air pressure of 150 Ibs. per square 
absolute, and back pressure of 10 Ibs. per square inch absolute, with \ .cut-off 

high-pressure cylinder and continuous expansion in low-pressure cylinder. Exponent 

expansion curve = 1.4 for compressed air, infinite receiver. 



272 ENGINEERING THERMODYNAMICS 

(a) From Eq. (332) 



which, on substituting values from above, gives for (m.e.p.) 63 Ibs. per sq. inch. 
Hence, the indicated horse-power =242. 
(6) From Eq. (334) 

1 Q '7 f\C\ *7 

Compressed air per hour per I.H.P. = - =- cu.ft., 

m.e.p. R c 

which, on substitution, gives 

13,750 .5 



(c) From Eq. (337) 



W 



H 



W L J z /JkV/l-^ 8 '^ /bk.pr. 
which gives 

5f -* i / O 



-=.294. 
\ 10 1 



and 

Wj/ + PTi=242I.H.P. 

Hence 

WW=56 I.H.P. 

and 

W L = 184 I.H.P. 

(d) From Eq. (330) 

(rec.pr.) = (in.pr.) ( ^ Y = ISO/ ^- r \ 1 ' 4 =57 Ibs. per sq.in. 



From Eq. (335) 

(rel.pr.)/, = (in.pr.)Z^ s , 



150x(.5) 14 =571bs. persq.in. 



WOEK OF PISTON ENGINES 273 

From Eq. (336) 



= 150 +21.85=6.85 Ibs. per sq.in. 

These values may be compared with those of Ex. 1, Section 9, which were for the 
same data with logarithmic expansion. 

Prob. 1. What will be the horse-power and steam used per hour by the follow- 
.ing engine under the conditions given? Cylinders 18 and 30x48 ins., speed 100 
R.P.M., initial pressure 150 Ibs. per square inch absolute, back-pressure 10 Ibs. per 
square inch absolute, steam continually dry. Cut-off at first in high-pressure and 
\ in low, and then ^ in each infinite receiver. 

Prob. 2. The very large receiver of a compound pumping engine is fitted with safety 
valve which is to be set to blow at 25 per cent above ordinary pressure. The cylinder 
ratio is 1 to 3.5, and cut-offs are f in high and | in low. If initial pressure is 125 
Ibs. per square inch gage, for what must valve be set? What vacuum must be carried 
in the condenser to have complete expansion in low-pressure cylinder? Superheated 
steam. 

Prob. 3. A compound engine is to be designed to work on superheated steam of 
125 Ibs. per square inch absolute, initial pressure and on an 18-inch vacuum. The 
load which it is to carry is 150 horse-power and piston speed is to be 500 ft. per 
minute at 200 R.P.M. Load is to be equally divided between cylinders and there is 
to be complete expansion in both cylinders. What must be cylinder sizes, and 
what cut-offs will be used for an infinite receiver? 

Prob. 4 V How will the economy of the two following engines compare? Each is 
14 and 20x24 ins., runs at 200 R.P.M. , on compressed air of 100 Ibs. per square 
inch gage pressure, and 15 Ibs. per square inch absolute exhaust pressure. Low-pres- 
sure cut-off of each is ^ and high pressure of one is I, the other, I. Infinite 
receivers. 

Prob. 5. A compound engine 12 and 18x24 ins. is running at 200 R.P.M. on 
superheated steam of 100 Ibs. per square inch absolute pressure and exhausting to a 
condenser in which pressure is 10 Ibs. per square inch absolute. The cut-off is I in 
high-pressure cylinder and | in low-pressure cylinder. Compare the power and steam 
consumption under this condition with corresponding values for wet steam under 
same conditions of pressure and cut-off and infinite receivers. 

Prob. 6. The initial pressure of an engine is 150 Ibs. per square inch absolute, the 
back pressure one atmosphere, the cylinder ratio 3. As operated, both cut-offs are at |. 
What will be the receiver pressure, high-pressure release pressure, and low-pressure 
release pressure? What will be the new values of each if (a) high-pressure cut-off is 
made i, (6) I, without change of anything else, (c) low pressure cut-off is made i, 
(d) f, without change of anything else? Infinite receiver, s = 1.3. 

Prob. 7. In the above problem for cut-off in each cylinder how will the release and 
receiver pressures change if (a) initial pressure is raised 25 per cent, (6) lowered 25 
per cent, (c) back pressure raised 25 per cent, (d) lowered 25 per cent? 

Prob. 8. How many pounds of initially dry steam per hour will be required to 
supply an 18-in. and 24x30-in. engine running at \ cut-off in each cylinder if speed 



274 



ENGINEERING THERMODYNAMICS 



be 200 R.P.M., initial pressure 100 Ibs. per square inch gage and back pressure 5 
Ibs. per square inch absolute? Expansion to be adiabatic and receiver infinite. 
NOTE: d for 100 Ibs. =.26. 

^ -Sup. Vol. 



INCOMPLE' [EXPANSION 




OVER EXPANSION V 

FIG. 84. Work of Expansive Fluid in Compound Engines with Finite Receiver, Zero Clear- 
ance. Cycle VII, Logarithmic Expansion; Cycle VIII, Exponential Expansion. 

11. Compound Engine with Finite Receiver. Logarithmic Law. No 
Clearance, Cycle VII. General Relations between Dimensions and Work 
when H.P. Exhaust and L.P. Admission are not Coincident. The diagrams, 
Fig. 84, while showing only two degrees of expansion, that of over and under 
in both cylinders, suffice for the derivation of equations applicable to all 
degrees in either cylinder. Volumes measured from the axis AL are those 



WORK OF PISTON ENGINES 



27,") 



occupied by the fluid in either cylinder alone, while fluid volumes entirely in the 
receiver, or partly in receiver, and in either cylinder at the same time are meas- 
ured from the axis A'L'. No confusion will result if all volumes represented by 
points be designated by the (V) with a subscript, and to these a constant rep- 
resenting the receiver volume be added when part of the fluid is in the receiver. 
Then, 



High-pressure cylinder work is 



(345) 



Low-pressure cylinder work is 
W L = P n O\og e 

Total work 



+P,Felog,^-P f F f . . . . (346) 



-P ( ,V . . (347) 



These expressions include some terms not known as initial data and may 
be reduced by the following relations, 



and 



Hence 



Dividing by the low-pressure cylinder displacement, V ff , the result will be 
the mean effective pressure referred to the low-pressure cylinder, 

(M.E.P. ref. to L.P.) 



276 



ENGINEERING THERMODYNAMICS 



A similar division but with the volume supplied, 
Work per cu.ft. supplied 



as the divisor, gives 






Also as in previous cases 
Cu.ft. supplied per hr. per I.H.P. 



(m.e.p. ref. to L.P.) 



(351) 



Of course, the weight per hour per I.H.P. follows from Eq. (351) by introduc- 
ing the density as a multiplier. 

While the last four equations can be used for the solution of problems, it is 
much better to transform them by introducing dimensional relations as in the 
previous cases developed. 

Let (rec.pr.)i= maximum receiver pressure P n , which is also the initial admis- 

sion pressure for the low-pressure cylinder; 

(rec.pr.) 2 = minimum receiver pressure P e , which is the terminal admission 
pressure for the low-pressure cylinder and that at which 
expansion begins there; 



" 



receiver volume 
high-pressure cyl. displ. 



____ H 

" ~ ' l ' 







Other symbols necessary are unchanged from the meaning imposed in Section (9). 
Substitution in Eqs. (348), (349), (350), and (351) gives the following set 
in a form for direct substitution of ordinary data: 



Work of cycle 

= U4(m.pr.)Z H D H 1 +log e -+log e - 



- (35) 



-log, (l+I 



- 144(bk.pr.)Z>z, (6) 



WORK OF PISTON ENGINES 



277 



(m.e.p. ref. to L.P.) 
= (m.pr.)f^ 1 1 +loge - 

tic 



- 



= (in.pr.) 



Work per cu.ft. supplied 

= 144(in.pr.)(l+log e -^+log e 



+ 1 + 



144(in.pr.) l+log e 



R L + 



Cu.ft. supplied per hr. per I.H.P. 

13,750 



(m.e.p. ref. to L.P.) R c 
13,750 1 



X 



(m.e.p. ref. to L.P.) R H Rc 



(6) 



(353) 



pr.)^ W 



(6) 



. (354) 



xf* (a) 



(355) 



It is desirable at this point to introduce a series of expressions fixing the 
relations between the dimensions, the cycle that may follow, and the fluctua- 
tions in the receiver pressure, and for the selection of cylinder and receiver 
dimensions for a required output of work and division of it between cylinders. 

In doing this it will be convenient to start with diagram points and finally 
substitute general symbols in each case. There will first be established the 
maximum and minimum receiver pressures and the fluctuations. 

Maximum receiver pressure 



=P IZe+o) = /py\ (Ve+ o\ _ p /T. , r,\ 




. (356) 



278 ENGINEERING THERMODYNAMICS 

Minimum receiver pressure 



(a) 



/^ 



(357) 



(in.pr.) 



Fluctuation in receiver] pressure = (P n Pe)=Pb-^> 

Z D Z 

.'. (rec.pr.)i (rec.pr.) 2 = (in.pr.) * " = (in.pr.)-j (a) 



(358) 



It is interesting to note that the minimum receiver pressure is exactly the same 
as the value of the constant-receiver pressure for infinite receiver, so that limit- 
ing the size of receiver does not affect the point E, but only raises point N higher, 
tending to throw more work on the L.P. cylinder for the same valve setting. 

The two release pressures P c and P/ can be evaluated as in the case of the 
infinite receiver, as both these points lie on the common expansion line, which 
is not at all affected by the receiver-pressure changes, and the values are the 
same as for the infinite receiver, and are here reproduced from Eqs. (309) and 
(310) with new numbers to make the set of equations complete: 



(rel.pr.)//=(in.pr.)~ (a) 
KH 

= (in.prr)Z* (6) 



(rel.pr.)z, = (in.pr.) -- (a) 



(359) 




(360) 



where R v is the ratio of maximum volume in the low- to the volume at cut-off 
in the high-pressure cylinder. 

Division of work between the cylinders cannot, as pointed out, be the same 
as for the infinite receiver, the tendency being to throw more work on the low 
as the receiver becomes smaller, assuming the cut-off to remain the same. As, 
therefore, equal division was obtainable in the case of infinite receiver with 



WORK OF PISTON ENGINES 



279 



equal cut-offs when the cylinder ratio was equal to the square root of initial 
over back pressure, it is evident that a finite receiver will require unequal cut- 
offs. As increase of low-pressure admission period or cut-off fraction lowers 
the receiver pressure and reduces the low-pressure work, it follows that with 
the finite receiver the low-pressure cut-off must be greater than the high for 
equal work division, and it is interesting to examine by analysis the ratio 
between them to determine if it should be constant or variable. 

For equal work division Eqs. (345) and (346) should be equal, hence by 
diagram points 



P,F 



(l+log, ) - 



r =PJ> 10g e 



+ P.V. log, - 



-P 



log, 



hence for equal division of work, the following relations must be satisfied : 



(log, R a - 



= 1 + 



Re 



/ 
\ 



bk.pr. 
in.pr. 



R H Rc-l. (361) 



It will be shown later that when expansion is complete in both cylinders 
and work equal that the high-pressure cylinder cut-off or the equivalent ratio 
of expansion bears a constant relation to that of the low, according to 

^ = 3, . (362) 

KL 

in which a is a constant depending only on the size of the receiver. It will 
also be shown that the cylinder ratio is a constant function of the initial and 
back pressures and the receiver volume for equal division of work, according to 

:^V, (363) 



in which (a) is the same constant as in Eq. (362). It is impo tant to know if 
these same values will also give equal division for this general case. Substi- 
tuting them in Eq. (361) 



280 



ENGINEERING THERMODYNAMICS 



Here there is only one variable, R L , the evaluation of which can be made by 
inspection, for if 



the equation will become 
2 log, a = (l+y) 
or 



which is a constant. 

1.001 



RECEIVER VOLUME EQUALS 
2/X H.P. DISPLACEMENT 




(364) 



.40 
High Pressure Cut Off 



,80 



.100 



FIG. 85. Diagram to Show Relation of High- and Low-Pressure Cut-offs for Equal Work in 
the Two Cylinders of a Finite-receiver Compound Engine with Zero Clearance 
and Logarithmic Law. 

As only one constant value of low-pressure ratio of expansion or cut-off 
satisfies the equation for equal division of work when there is a fixed ratio 
between the values for high and low, that necessary for equal division with 
complete expansion in both, it is evident that equal division of work between 
the two cylinders cannot be maintained at all values of cut-off by fixing the 
ratio between them. As the relation between these cut-offs is a matter of some 
interest and as it cannot be derived by a solution of the general equation it is 
given by the curve, Fig. 85, to scale, the points of which were calculated. 



WORK OF PISTON ENGINES 



281 



A special case of this cycle of sufficient importance to warrant derivation 
of equations because of the simplicity of their form and consequent value 
in estimating when exact solutions of a particular problem are impossible, 
is the case of complete and perfect expansion in both cylinders. For it the 
following equations of condition hold, referring to Fig. 84, 



il 




INDICATOR CARDS OF EQUAL 
BASE AND HEIGHT 



!H 



IG. 86. Special Case of Cycles VII and VIII Complete Expansion, in both Cylinders of the 
Finite Receiver Compound Engine. Zero Clearance. 

which when fulfilled yield the diagram, Fig. 86. These equations of conditions 
are equivalent to fixing the cut-off in both high- and low-pressure cylinders, 
and the volume of the high- with respect to the low-pressure volume . Accordingly, 



(365) 



Also for the low-pressure cylinder the cut-off volume must equal the whole 
ligh-pressure volume, or D H = Z L D L . Therefore, 






= ' (a) 
He 



(366) 



2X12 



ENGINEERING THERMODYNAMICS 



Substituting these equations of condition in the characteristic set Eqs. (352), 
(353), (354), and (355), there results the following for most economical operation: 



Work of cycle 
W 



log, 



-144(bk.pr.)I> i 



(a) 



144(in.pr.)Z)z, ^ = 144(in.pr.)D i 



/in.pr. \ 
Vbk.prJ 



R 



(367) 



(m.e.p. ref. to L.P.) = 



i /in.pr. \ 

/in.pr. \ ,. ,. ge \bk.pr./ ,. , 
(bk.pr.) log, ^ J = (m.pr.) . = (m.pr.) 



e - /0 ^o\ 
-- . (368) 



W 

Work per cu.ft. supplied = -^fr- 



(bk.prODz, . /in.pr. \ .... V1 /in.pr. 

144 , , loge , , =144(m.pr.) lofotri 

p /bk.pr A 5 \bk.pr. / 5 Vbk.pr. 

J \in.pr. / H 



. (369 
Cu.ft. supplied per hr. per I.H.P. 

13,750 x f bk -P r A- 13,750 1 , 

(m.e.p. ref. to L. P.) * \in.pr. / (m.e.p. ref. to L. P.) *R V ' 

For this special case of best economy the receiver and release pressures 
of course, have special values obtained by substituting the equations c , 
condition Eqs. (365) and (366), in Eqs. (356), (357), (358), (359), (360). 



(rec-p,), = (in.pr.) + = (in.pr.) + 



WOKK OF PISTON ENGINES 



283 



Therefore 



r> 

.pr.)-~; 



(373) 
(374) 
(375) 



These last two expressions might have been set down at once, but are 
orked out as checks on the previous equations. 
For equal division of work in this special case the general Eq. (361) becomes 



(rec.pr.)i-(rec.pr.) 2 

(rel.pr.)*=(in.pr.) =(bk.prO#c=(rec.pr.) 2 ; 
K H 



Therefore 



(376) 



'his term, a, has already been used in previous discussions of equality of 
prk, while the derivation of its value has not been made up to this point. 

This indicates that ratio of cut-offs or individual ratios of expansion is a 
inction of the receiver size for equal division of work. 
From Eq. (376) the cylinder ratio can be found in terms of a, and the ratio 
expansion. Referring to Fig. 86, 



2 



V e 







ice the cylinder ratio is equal to a constant depending on the receiver size, 
'Ultiplied by the value for the infinite receiver, i.e., the square root of the initial 
by back pressure. 



284 ENGINEERING THERMODYNAMICS 

The high-pressure cylinder ratio of expansion is 




. . . (378) 



and the corresponding value for the low-pressure cylinder is 

V, 1 Vl /5nr7 



, . . . (379) 



For convenience in calculation Table, XII of values of a and a 2 is added foi 
various size of receivers. 

TABLE XII. 



Receiver Vol. 


[(l+y)log e (l + ^-)-l] 


a 2 


H.P. Cyl. Disp. l 






.5 


1.915 


2.64 


.75 


1.624 


3.67 


1.0 


1.474 


2.17 


1.5 


1.322 


.75 


2.0 


1.243 


.55 


2.5 


1.198 


.437 


3.0 


1.164 


.359 


4.0 


1.1223 


.262 


5.0 


1.0973 


1.204 


7.0 


1.0690 


1.143 


10.0 


1.0478 


1.098 


14.0 


1.0366 


1.068 


20.0 


1.0228 


1.046 


Infinite 


1.0 


1.0 



At the end of this chapter there is presented a chart which gives the relatio] 
between cylinder and receiver volumes, cylinder ratio, and high- and low-pressur 
cut-offs graphically. 

The corresponding values of maximum and minimum receiver pressun 
for equal division of work for this case of best economy are 



(rec-pr.), - (rec.pr.) 2 = V ^^'\ 



(382 



WORK OF PISTON ENGINES 285 

Example 1. Method of calculating Diagram, Fig. 84. 
Assumed data for case A: 

p a =p b = 120 Ibs. per sq.in. abs. V a = V n = V m =0 cu.ft. 
P m =P = 10 Ibs. per sq.in. abs. V* = A cu.ft. 

V e = l cu.it. 

= 1.2cu.ft. 

V e = .Scu.it. 
To find point (7: 



.8 
To obtain point E: 



To obtain point D: 

Pe(V e - 

To obtain point N: 



P e =P 6 = . =48 Ibs. per sq. inch. 

Ve 1 



48 v9 9 
Pe(F,+0)=P d (Fc+0) or P d =^^=53 Ibs. per sq. inch. 



48 v2 2 
n )=P e (Fe+0) or P n =-^-=88 Ibs. per sq. inch. 



To obtain point F: 






48x1 . . , 

=24 Ibs. per sq. inch. 



Example 2. Find (a) the horse-power, (6) steam used per hour, (c) the release 
and receiver pressures for a 12- and 18x24-in. engine with receiver twice as large as 
the low-pressure cylinder when the initial pressure is 150 Ibs. per square inch absolute, 
back pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., and cut-offs | in 
aigli-pressure and such a value in the low pressure as to give complete expansion, and 
learances zero. 

(a) From Eq. (353) 



jm.e.p.) = (in.pr.) - 1 1 +loge RH -flog* RL 



which on substituting the above values gives 

A. K\ / \\ 

-10 = 73.3 Ibs. 



le 



nce I.H.P.=282. 
(6) From Eq. (355) we have 

fu.ft. steam per hour per horse-power = ! - - X p p =-~~^- X =41.7, 

(m.e.p.) Katie 



286 



ENGINEERING THERMODYNAMICS 



(c) From Eqs. (356) and (357) for maximum and minimum receiver pressures 
respectively : 

+ R L \ and 
RnRcl 



maximum receiver pressure = 



I 2 25 

-+--- =91.5 Ibs. per sq. inch. 



2.25 



minimum 



receiver pressure = 150 X^r~-^= 75 Ibs. per sq. inch. 



From Eqs. (359) and (360) for release pressures 
(in.p,)Z and 






high pressure cylinder release pressure = 1 50 X. 5 =75 Ibs., per sq. inch. 






low pressure cylinder release pressure = -=33.9 Ibs. per sq. inch. 

'' 



These results may be compared with those of Example 1 of Sections 9 and 10, whicl 
are derived for same engine, with data to fit the special cycle described in the particular] 
section. 

NOTE: In all the following problems clearance is to be neglected. 

Prob. 1. A 12- and 18x24-in. engine has a receiver equal to 5 times the volume of! 
the high-pressure cylinder. It is running on an initial pressure of 150 Ibs. per square 
inch gage and exhausts to the atmosphere. It has a speed of 150 R.P.M. and the cut-offs 
are -& and } in high- and low-pressure cylinders respectively. What is the horse-poweij 
and the steam used in cubic feet per hour? 

Prob. 2. What will be the release pressures, and variation of receiver pressure foil 
an engine in which the cylinder ratio is 3, cut-offs | and |, in high and low, initial pres- 
sure is 100 Ibs. per square inch absolute, and receiver 2 times low-pressure cylindei] 
volume? 

Prob. 3. Show whether or not the following engine will develope equal cylindei 
work for the conditions given. Cylinder diameters, 15 and 22 in., initial pressure 
Ibs. per square inch gage, back pressure 10 Ibs. per square inch absolute, cut-offs ] 
and f , receiver volume 4 times high-pressure cylinder, strokes equal. 

Prob. 4. For the same conditions as above, what low-pressure cut-off would giv<| 
equal work? 

Prob. 6. What will be the most economical load for a 16- and 24x30-in. engii 
running at 125 R.P.M. on 150 Ibs. per square inch absolute initial pressure and atj 
mospheric backpressure? What will be the economy at this load? 

Prob. 6. What will be the release and receiver pressures for the above engine i| 
the receiver has a volume of 15 cu.ft.? 



WORK OF PISTON ENGINES 287 

Prob. 7. Find the cut-offs and cylinder ratio for equal work division and complete 
expansion when initial pressure is 150 Ibs. per square inch absolute and back 
pressure is 10 Ibs. per square inch absolute, receiver four H.P. volumes. 

Prob. 8. Will a 14- and 20x20-in. engine, with a receiver volume equal to 5 tunes 
:he, H.P. cylinder and running on I cut-off on the high-pressure cylinder and I cut-off 
MI the low, with steam pressure of 100 Ibs. per square inch gage and back pressure 
)f 5 Ibs. per square inch absolute, have complete expansion and equal work distri- 
Dution? If not, what changes must be made in the cut-off or initial pressure? 

Prob. 9. What must be the size of an engine to give 200 1. H.P. at 150 R.P.M. on an 
nitial steam pressure of 150 Ibs. per square inch absolute, and 10 Ibs. per square inch 
ibsolute back pressure, if the piston speed is limited to 450 ft. per minute and complete 
expansion and equal work distribution is required? Receiver is to be 6 times the volume 
rf high-pressure cylinder and H.P. stroke equal to diameter. 

12. Compound Engine with Finite Receiver. Exponential Law, No 
Clearance. Cycle VIII. General Relations between Pressures, Dimensions, 
and Work, when High Pressure Exhaust and Low-pressure Admission are 
Independent. The diagram Fig. 84 may be used to represent this cycle, as well 
as cycle VII, by conceiving a slight change in the slope of the expansion and 
receiver lines. Using the same symbols as those of the preceding section, 
and the expression for work as found in Section 7, Chapter I, 



-^r^k-l^-'] } 



nd the last term in the equation for W H within the bracket may therefore be 

written 

Cm.pr.)Z, (y\ (Z,\ /_ y_ +l \ ^ pLY 
-~ l \y+l) 



- 

Rcz 



2SS ENGINEERING THERMODYNAMICS 

and hence by simplifying the first two terms also, 



Work of the low-pressure cylinder may be expressed in terms of pressure 
and volumes at N, E, and G, but it is convenient to use instead of the pressure 
at N or at E, its equivalent in terms of the point B. The pressure at N is 



and when multiplied by the receiver volume yDn, it becomes 



At E the product of pressure and volume is 
(rec.pr.) 2 



Using these quantities, the following equation gives the work of the low- 
pressure cylinder: 



and the total work is, by adding (W H ) and (W L ), 



This Eq. (385) is the general expression for work of the zero clearance com- 
pound engine with exponential expansion, no clearance, and finite receiver, 
From this the following expressions are derived : 

(m.e.p. ref. to L.P.) 

-i 







WORK OF PISTON ENGINES 289 

Vork per cu.ft. supplied is 

_144^I.-Z a .-'+^V~ 1 p- + lY IY-JLV- 1 
-! 1 \v) \RcZ,7 V l\y+l/ 

.pr.). . (387) 



}u.ft. supplied per hr. per I.H.P. 



13,750 



(m.e.p. ref. to L. P.) R c ' 



(388) 
(389) 



; .... (390) 

*=(in.pr.)Zj/; ..... ..... (391) 

V ..... (392) 



If work is equally divided between the cylinders, W H , Eq. (383), and W L , 
(384), will become equal, hence 



._, 

- 



. 

(in.pr.) Z ff 

This equation shows conditions to be fulfilled in order that an equal division 
work may be obtained. It does not yield directly to a general solution. 
When expansion is complete in both cylinders, 



1 /bk.prA (Z H \> 

^- and -. ) = l 7^- ) . 
Rc \in.pr. / \Rc/ 



Rc 

Introducing these values in the general expression Eq. (385) for work of 
jiis cycle, it may be reduced to the following: 



..... (394) 



From which are obtained 

(m.e.p. ref. to L. P.) = (in.pr.)-p- 



~ T [l-^y" j 



s f (Zn\'~ 1 ~\ 
r ork per cu.ft. supplied = 144(in.pr. -377 1~ l"p~ ) 



. (395) 
(396) 






290 ENGINEERING THERMODYNAMICS 

13,750 _ 
Cu.ft. supplied per hr. per I.H.P. = (m e p ref to L p } 

13,750 



(WESLV. (397 

. P.) \in.pr. / 



(m.e.p. ref. to L 

If work is equally divided and complete expansion is maintained in botl 
cylinders Eq. (381) becomes 



which may be simplified to the form, 

-s. ... (398 



where Rv is the ratio of maximum low-pressure volume, to the high-pressui 
volume at cut-off, 



hence 



and the value of Rv may be found from original data, 



Er-Mc L ) (39 

Vbk.pr./ 

Eq. (398) may easily be solved for Z H , from which the required cylind 
ratio may be found by, 

Rc=Z H R v . ......... (40( 

This is the cylinder ratio which gives equal work in the two cylinders am 
complete expansion in both, when used with the value found for the higl 
pressure cut-off Z H , the assumed initial and back pressures, and the assume 
raio, y, of receiver volume to high-pressure displacement. 



WORK OF PISTON ENGINES 



291 



Example. Find (a) the horse-power, (6) steam used per hour, (c) the release 
nd receiver pressures of a 12- and 18x24-in. engine, with a receiver twice as large as 
ic low-pressure cylinder when the initial pressure is 150 Ibs. per square inch absolute, 
ack pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., and cut-offs in the 
igh and such a value in the low as to give complete expansion. Exponent for ex- 
ansion curve = 1.4. 

(a) From Eq. (386) 



m.e.p., 



yhich, on substituting above values, gives 
50 .5 [ . '_ / .5 \ - 4 / 4.5 




'fence 



(m.e.p.) =57.5 Ibs. per sq.in., 
I.H.P.=221. 



(b) From Eq. (388) 



r, , 13,750 Z H 

Oubic reet 01 steam per hour per horse-power = =. 

m.e.p. Re 



13,750 .5 



lence total pounds per hour will be 

53.2 X221X. 332 =3910. 

From Eqs. (389) to (392): 

J Z H \ s (,RcZ L \ s 
(rec.pr.)i = (m.pr.) (--) (l +--) , 



(7 \ s 
T I 



(rel.pr.)^ = (m.pr.)Z H s , 






292 ENGINEERING THERMODYNAMICS 

These, on substitution of the proper numerical values, become: 

(g\ 1.4 / j \ 1.4 

~j f Xl + ) =75 Ibs. per sq. inch, 

(rec.pr.) 2 = 150x(.5) 1 - 4 =57 Ibs.. 



=57 Ibs, 
(rel.pr.)z, =57 X = 32 -! lbs - " 



NOTE: In all the following problems clearance is assumed to be zero. 

Prob. 1. A 12x18x24 in. engine is running on superheated steam of 150 Ibs. per 
square inch absolute pressure, and exhausts to the atmosphere. If the speed is 100 
R.P.M., high-pressure cut-off |, low pressure cut-off , and receiver volume 10 
cu.ft., what horse-power will be developed and what steam used per hour? 

Prob. 2. What would be the effect on the power and the economy of (a) changing i 
to wet steam in the above? (6) to compressed air? 

Prob. 3. What would be the receiver and the release pressures for each case? 

Prob. 4. Will there be equal work distribution between the two cylinders? 

Prob. 5. It is desired to obtain complete expansion in a 14x22x36-in. engine 
running on fluid which gives a value for s of 1.2. Initial pressure is 100 Ibs. per 
square inch gage, and back pressure 5 Ibs. per square inch absolute. What must be 
the cut-offs and what power will be developed at 500 ft. piston speed? Receiver = 
XH.P. volume. 

Prob. 6. How large must the receiver be for the above engine in order that the 
pressure in it shall not fluctuate more than 5 Ibs. per sq. inch? 

Prob. 7. An engine is to run on steam which will give a value of s = l.l, and to 
develope 500 horse-power at 100 R.P.M. Piston speed is not to exceed 500 ft. per 
minute. Steam pressure, 150 Ibs. per square inch absolute, back pressure, 5 Ibs. per 
square inch absolute. Complete expansion and equal work distribution, for this load are 
to be accomplished. What will be the cylinder sizes and the high-pressure cut-off if the 
receiver is to be 3 times the high-pressure cylinder volume? 

Prob. 8. What will be the steam used per hour by the engine of Prob. 7, and 
what will be the variation in the receiver pressure? 

Prob. 9. If the high-pressure cut-off were halved, how would the power and 
economy be affected? 

13. Compound Engine without Receiver, Logarithmic Law. No Clear- 
ance, Cycle IX. General Relations between Dimensions and Work when 
High-Pressure Exhaust and Low-Pressure Admission are Coincident. Such 
a peculiar case as this admits of but little modification of the cycle compared 
with the receiver cases, b: cause the low-pressure expansion is necessarily a direct 
continuation of the high pressure without any possible break. There can be no 
over-expansion in the high nor can expansion there be incomplete, as there is, 
properly speaking, no back pressure with which to compare the high-pressure 
cylinder terminal pressure. There may, however, be over and incomplete 
expansion in the low-pressure cylinder. It might appear that the high-pres- 
sure cylinder negative work was equal to the low-pressure admission work, as 
each is represented by the area below DC, Fig. 87A, but this is not the case, since 



WORK OF PISTON ENGINES 



293 



le diagram is drawn to two different scales of volumes, showing the pressure- 
jroke relation between high and low. This is apparent from the diagram, Fig. 
1C showing fluid volumes in each cylinder to a single scale on which ABCD 
; the work done in the high-pressure cylinder, ABD'EF the whole work, whence 

There is, of course, 
The cycle, 



)CD'EF is the part done in the low-pressure cylinder. 

o low-pressure cut-off or even admission as ordinarily considered. 

j far as the work to be done is concerned, is the same as for a simple engine, 



H.P.Cyl. Vols. 




INDIVIDUAL CYLINDER WORK SHOWN TO SAME 
SCALE OF PRESSURE AND VOLUME 



N K L 

HIGH AND LOW PRESSURE CYLINDER DIAGRAMS PLOTTED TO SAME AXIS 



'IG. 87. Work of Expansion in the No-receiver Compound Engine, Zero Clearance, Cycle 
IX, Logarithmic Expansion, Cycle X, Exponential Coincident Piston Movement. 

nd the only reason for introducing formulas for overall work, work per cubic 
ypt supplied, (m.e.p. referred to low), and fluid consumption, is to put them into 
orm for immediate substitution of dimensional relations. Because of the absence 
f cut-off in the low, the distribution of work between high and low will 
lepend solely on the cylinder ratio and high-pressure cut-off, for, the earlier the 
agh-pressure cut-off, and the larger the high-pressure cylinder, the greater 
lie fraction of the total work that will be done there, as there is only a fixed 
mount available, and the less there will be left to be done in the low 



294 ENGINEERING THERMODYNAMICS 

The diagrams of the two . cylinders are plotted to combined axes in 
Fig. 87 D. The points Q and R at equal heights. KN is the L. P. displace- 
ment, and KG that of the H.P. It has been shown in Section 8, that 
the expansion lines CD and C'D' may be plotted to the axes LN and LXM, 
the point X being the intersection of NQ and KR extended, and that the distance 

T-N RC 



= ^^. DH= ^_ =Dif _l__. f . _ i (4Q2) 




Hence the work area under CD is 



, I J- *- XV "i A A / 1 \ *-** H 1 -*--/ / 

GL DL DH * D H 



but 



hence 



( 1 1 

tf = 144(in.pr.)ZtfDtf| l-flog e ^ 

^H RC~ 



Rc\. . (404) 



Again the work area under C'D' is 

=^=144(rel.pr.)^- L " W e -~ t 

If I *'/)_ fi & e 7^ ' 



hence 

TFi / = 144(in.pr.)Z // D // ^^--jlog e ^ c -144(bk.pr.) J D L , . (4C5) 
and the total work, 

W = 144(in.pr.)2W>* J 1-flog, ^~ ^ loge R c 



WOEK OF PISTON ENGINES 



295 



But 
and 
so that 



/ Re _i\_i 
\Rc-l Rc-l) ' 



loge 



i **O i r> 

= iOg e =- = lOge HVJ 



"H 



. . (406) 



which shows by its similarity to the work of the simple engine that, as before 
stated, the total work is the same for this cycle as if the entire expansion were 
made to take place in a single cylinder. 

This same result could have been attained in another way sufficiently 

; interesting to warrant setting it down. Since the low-pressure work is repre- 
sented truly to scale by C'D'EF, Fig. 87C, the mean effective pressure of the 
low-pressure cylinder is given by the area divided by V e . By contracting all 

'volumes proportionately, C'D' takes the position CD' and C'F the position 
CF f , hence 

area CD'EF 



V.-Vf. 



-P. 



represents the mean effective pressure in the low-pressure cylinder just as 
truly. Therefore, 

" ,.' , iare&CD'EF D \ Tr 
L.P. cylinder work= ^ ^ P* V e 



As the high-pressure work is (total low), 
H.P. cylinder work = P 6 F/l+log e ^ \ -P e V e -. 

. / f ; 



r c ^ +P e F< 



F 



V e 



Introducing symbols 
L.P. cylinder 

H.P. cylinder 



r.)D I ,. (407) 
|^--^- 1 \og 6 Rc\ 

-^loge^c. . (408) 



iich check with Eqs. (404) and (405), 



296 



ENGINEERING THERMODYNAMICS 



Dividing the total work by the low-pressure cylinder volume and the high- 
pressure admission volume in turn, 



(m.e.p. ref. to L.P.) = (in.prO 



? - (bk.pr.) 



(a) 



- (bk.pr.) (6) 



. . (409) 



Re 



Work per cu.ft. supplied = 144(in.pr.)Z#( l+log e -~\ - (bk.pr .)R C (a) 



. (410J 



Cu.ft. supplied per hr. per I.H.P. = ^^f^ LJ > ^ (a) 



13,750 1_ 

(m.e.p. ref. to L.P.) R H Rc 



(&) 



. . (411 



For equal division of work there can obviously be only one setting of th 
high-pressure cut-off for a given cylinder ratio and any change of load to be me 
by a change of initial pressure or of high-pressure cut-off will necessarily unbalanc 
the work. Equating the high-pressure and low-pressure work expressions 
Eqs. (404) and (405), 



or 



p - = 

Kc~ 1 

(bk.pr.) 



Re 

= ^ - = 

Kc 1 
Rc+l 



/bk.prA R c 

-( r - 3? 

\in.pr. / L H 



Another relation exists between Z H and R c , namely, that 

= Rv 

f7 ~ r> > 

AH 'Me 
where Rv is the ratio of volumetric expansion. Then 



but 



hence 



WORK OF PISTON ENGINES 



297 



With this formula it is possible to find the necessary ratio of cylinder 
displacements for given initial and back pressures and for given ratio of 
expansion R v . 

For convenience in solving this, a curve is given in Fig. 88 to find value 

2R 

c 

of Re when Rc R c ~ l has been found. 



37 



? 

3 






























^ 


x^ 


"*^ 




























x^ 


-^ 




























. 


x^ 


' 




























x 


^ 






























. ^ 


/ 
































/ 


r 
































X 
































, 


/ 


































/ 


































, 25 50 75 100 2Rc 125 150 
"Values of (Re) "*& 



i. 88. Curve to Show Relation between Values of RE and (RE) R C l for Use in Solving 
Eq. (412), Giving Cylinder Ratio in Terms of Ratio of Expansion for the No-receiver 
Compound Engine without Clearance. 

The complete expansion case of this cycle results from the condition 


P d =Pe or (r6l.pr.)ji=(bk.pr.) or R v 



rhich when applied to Fig. 87, transforms the diagrams to the form Fig. 89. 
It also follows that 

(bk.pr.)z, = (m.pT.)Z H D a 
and 



Rc_ /in.pr. 
~Z~ H ~ \bk.pr. 



These conditions will, of course, reduce the total work Eq. (406) to the 
common value for all cycles with logarithmic expansion and likewise those 
for mean effective pressure, work per cubic foot supplied, and consumption. 
For the equal division of work under this condition, Eq. (412), becomes 



(413) 



n.pr. 



= i and R may represent ratio of expansion or ratio of 



298 



ENGINEEEING THERMODYNAMICS 



initial to back pressures, these being equal. Fig. 90 gives a curve showing the 
relation between cylinder ratio and ratio of expansion established by the above 
condition. 



H J>. Cyl. vols. 




54321 V 
L.P. Cyl. Vols. 



\ 



\ 




FIG. 89. Special Case of Cycles IX and X. Complete Expansion in both Cylinders of the 
No-Receiver Compound Engine, Zero Clearnace. 

Example 1. Method of calculating Diagram, Fig. 87. 

A. As described in the text this diagram is drawn to two-volume scales, so tha 
there may be two volumes for one point. 



WORK OF PISTON ENGINES 



299 



Assumed data: 

p a =p b = 120 Ibs. per sq.in. abs. 
p e =p f = 10 Ibs. per sq.in. abs. 



To locate point C: 



To locate point D: 



V a = V c = V d = V g = V f =0 cu.ft. 
F& = 1 cu.ft. 
F c =2cu.ft. 



p 120X1 A01U 

PC =-r 7 = - =60 Ibs. per sq.in. 

r c & 



p PcV c 60X2 

Pa = ;; = - =24 Ibs. per sq.in. 

V d O 



To locate intermediate points from C to D. The volume at any intermediate 
tint is (the volume of low-pressure cylinder up to that point) +( volume of high- 



Values of Re 

M CO O1 <] <O 








































* 


. 









































_^ 


.--- 


*" 






































^ 


"-" 








































^ 


"" 










































^ 


** 










































X" 












































/- 












































/ 


/ 












































/\ 














































5 10 15 20 

Values of R 



EG. 90. Curve to Show Relation between Values of RC, the Cylinder Ratio, and Rihe Ratio 
of Initial to Back Pressure for Complete Expansion in the No-receiver Compound 
Engine without Clearance (Eq. (413).) 

pressure cylinder from that point to end of stroke), e.g., at I stroke the volume in 
low, is .75x5, and the volume in the high is .25x2, or total 4.25, and the pressure 
at that point is found by the PV relation as above. 



B. Assumed data: 



P a =p b =120 Ibs. per sq.in. abs. 
P e =Pj= 10 Ibs. per sq.in. abs. 



F a = F/=0 cu.ft. 

F d = F e =5cu.ft. 

F 6 = l cu.ft. 



To locate point D: 



_ P 6 F & 120X1 . ., 
P d --2-- =24 Ibs. per sq.in. 
Va 5 



Intermediate points from B to D found by assuming volumes and computing 
>ressures from the PV relation as above. 

C. Figure ABCD constructed as in A. Figure C'D'EF is figure CDEF of A to the 
ame pressure scale but to a volume scale 2.5 times as large. 

D. Figures constructed as in C. 



300 ENGINEERING THERMODYNAMICS 

To draw indicator cards. The volume and pressure scales are chosen and from 
diagram A, a distance AB is laid off to the volume scale, AD is then laid off equal to 
AD of diagram A to the pressure scale. Point C is located to these scales and joined 
to B and D by drawing curves through the intermediate points plotted from the PV 
diagram to the scales of the card. For the low-pressure card EF is laid off to the 
volume scale, and FC' and ED' to pressure scale. C' and D' are then joined in same 
manner as C and D for high-pressure card. 

Example 2. Find (a) the horse-power, and (6) steam used per hour for a 12 X 18 X 24 
in. engine with no clearance when initial pressure is 150 Ibs. per square inch absolute, 
back pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., and cut-off in 
the high-pressure cylinder is |, there being no receiver. 

(o) From Eq. (409) we have 


1 

(m.e.p.)=(in.pr.) p p [1 +log e (###c)] -(bk.pr.), 



-10=50 Ibs. sq.in. 

hence 

I.H.P.=192. 

(b) From Eq. (411) we have 

13 750 1 

Cubic feet of steam per I.H.P. per hour =7 X -^-, 

(m.e.p.) R H ttc 

_13,750 _^_ 
50 X 2X2.25 

hence the weight of steam used per hour will be 

61 .2 X .332 X 192 = 3890 pounds. 

Example 3. What will be the cylinder ratio and the high-pressure cut-off to give 
equal work distribution for a ratio of expansion of 6, an initial pressure of 150 Ibs. per 
square inch absolute and back pressure of 10 Ibs. per square inch absolute? 

Ratio of back to initial pressures is .067 and 



hence from Eq. (412) 

2R C 

loge 7 ^ --1.40, 
or 

2R C 



Rc R c -i =24.36, 
and from Fig. 88 



7? *? & 

From the relation Z# = c = high-pressure cut off = r = .446. 
liv o 



WORK OF PISTON ENGINES 301 

Prob. 1. A compound locomotive has no receiver and runs on an initial pressure of 
175 Ibs. per square inch gage and atmospheric exhaust. The cylinders are 18 and 
50 X42 in. The steam pressure may be varied, as may also the cut-off to a limited 
legree. For a speed of 200 R.P.M. and a constant cut-off of f , find how the power 
ivill vary with initial pressure and for constant initial pressure equal to boiler pres- 
sure show how the power at the same speed will vary from % cut-off to full stroke. 

Prob. 2. Show how the steam used per horse-power hour will vary in above 
problem. 

NOTE: d for 175 Ibs. =.416. 

Prob. 3. With the cut-off at f, what should the initial pressure be to give equal 
work distribution? 

Prob. 4. With full boiler pressure and J cut-off what would be terminal pressure 
in the low-pressure cylinder? 

Prob. 5. What must be size of cylinders for a tandem compound engine with 
negligable receiver volume to run at 125 R.P.M. with complete expansion and equal 
|work distribution on an initial pressure of 125 Ibs. per square inch gage and a back 
pressure of 5 Ibs. per square inch absolute, when carrying a load of 500 horse-power, 
'the piston speed to be less than 500 ft. per minute? 

Prob. 6. What will be the steam used by the above engine in pounds per hour? 

NOTE: 8 for 125 Ibs. =.311. 

Prob. 7. A builder gives following data for a tandem compound steam engine. 
Check the horse-power and see if the work is equally divided at the rated load. 
Cylinders 10 ins. and 17^X15 ins., initial pressure 125 Ibs., speed 250 R.P.M., horse- 
power 155. Neglect the receiver volume. 

Prob. 8. Another manufacturer gives for his engine the following, check this: 
Cylinders 20 and 32x18 ins., initial pressure 100 Ibs., atmospheric exhaust, speed 
200 R.P.M., horse-power 400. Neglect the receiver volume. 

14. Compound Engine without Receiver, Exponential Law, Cycle X. 
General Relations between Dimensions and Work when High-pressure 
Exhaust and Low-pressure Admission are Coincident. Referring to Fig. 
87 D it is desirable first to evaluate the work areas CDKG and C'D'NK. As 
before, 



t 



= D H ~-r and 



hence 



(rel.pr.)#=(in.pr.)Zj/, 




?[!_ /J-V- 1 
-l \R C ) 



302 ENGINEERING THERMODYNAMICS 



whence 



.. - .r.)^. -. . (415) 

81 L Kc 1 J 

It is to be expected that the sum of high- and low-pressure work will be 
of a form similar to that which would be obtained if all work were performed 
in a single cylinder of a displacement equal to that of the low pressure, adding, 



W = WH+W L = 144(in.pr.) 



DHZ H \ _ 7 ,., 

L-I L V Rc-i 

-144(bk.pr.)Dz, 



r /iv-ni 

+ZJST- |l - ^-j II- 144(bk.pr.)I>z, 



7 1 



whence, substituthig D^=IT- 



r -)^; (416) 

f rft TTM (m.pr.) Z g f /M 8 " 1 ! 

: ^^~5;L S -(^J J-(bk.pr.) ..... (417) 

Work per cu.ft. supplied = 144 { m *"| ') L_ (i) 8 " 1 -144(bk.pr.) (418) 

L \F/ J Z ff 

Cu.ft. supplied per hr. per I.H.P. = 13 ' 750 ?? 

(m.e.p. ref. to L.P.) R c 

13,750 

(mie.p. ref. to L.Pj R v ' ' ^ 41 

Conditions for equal division of work between high- and low-pressure 
cylinders may be obtained by equating Eqs. (414) and (415). 

s-Z H -*-Z H -i 



= Z-iR c \- A^_l t**T.\R c 

L Rc-1 J WprJ^ 5 " 1 )* 



WORK OF PISTON ENGINES 303 

Rearranging 



The last term in the first member of this equation may be expressed as 

/bk.prA,, , 
( }Rv(s 1) 
\in.pr. ) 



and the relation 

Re 



IH= 



xists between Z H and Re, hence, making these substitutions, 

^'W(-DI- (420) 



which is not a simple relation, but can be solved by trial. 

The assumption of complete expansion in the low-pressure cylinder (it is 

(ways complete in high, for this cycle), leads to this following relations: 
/in.pr. \ _ f 
I in I K>v t 
\bk.pr./ 

I hence 

144(bk.pr.)D L = 144(in.pr.)^D^, 
and from Eq. (414), 

TF*=144(in r )^ \s- ( V' 1 - -'] 

o L \ ' / " * -1 

but 

R c Rys-l s-1 1 /bk.pr.Xs 

-p-=Rv, and ^- ^p~r = ^ s -v and ' ^~ = \in~V/ 

s-l 

... (421) 



(m .e.p. ref. to L.PO^in.pr.^l-. . . (422) 



304 



ENGINEERING THERMODYNAMICS 



The expression for equality of work Eq. (420) becomes, for this case of complete 
expansion, 



Rcl 
Rc-l 



. . . (423) 



by which it is not difficult to find the ratio of expansion R v , which gives equality 
of work for given values of s, and Re, the cylinder ratio. Values for R v for 
various values of R c and s are given by the curves of Fig. 91. 



10 




* 

s 

I 5 



1.5 



1.4 



Values of b 
1.3 12 



1.1 






10 

Values of Rv 



15 



FIG. 91. Curves to Show Relation between R c the Cylinder Ratio, and H v the Ratio of 
Expansion, for Various Values of (s), Applied to the No-receiver Compound Engine 
without Clearance, when the Expansion is not Logarithmic. 

Example 1. Find (a) the horse-power, and (6) the steam used per hour for a 
12- and 18x24-in. engine with no receiver when the initial pressure is 150 Ibs. per 
square inch absolute, back pressure 10 Ibs. per square inch absolute, speed 125 
R.P.M., cut-off in high-pressure cylinder is J, there being no receiver and steam 
having expansion, such that s = 1.3. 

From Eq. (417) 



which, on substituting the above values, becomes 



sq.in. 



hence the indicated horse-power =243. 



WORK OF PISTON ENGINES 305 

(6) From Eq. (419) the steam used per hour in cu.ft. per horse-power is 

. 
' 

13,7502^ 
m.e.p. Re 

which, for the data given above, becomes 

13,750 .5 



or pounds per hour total, is, 48.2 X 243 X. 332 =3880. 

Example 2. What will be the high-pressure cut-off and cylinder ratio to give 
equal work distribution and complete expansion for an initial pressure of 150 Ibs. per 
square inch absolute, and back pressure of 10 Ibs. per square inch absolute? 

From relation # F s = (r^ -), #F = 6.9 and from this, by the curve of Fig. 91, 
\DK.pr./. 



For complete expansion 



Re 5.4 _ 
-~~~ 



Prob. 1. A tandem compound engine without receiver has cylinders 18- and 

X42-ins. and runs at 200 R.P.M. What will be the horse-power developed at 
ds speed if the initial pressure is 175 Ibs. per square inch gage, back pressure 
itmosphere, high-pressure cut-off ^, and s has a value of (a) 1.1, (6) 1.3? Compare 
the results with Prob. 1 of Sec. 13. 

Prob. 2. What will be the weight of steam used per horse-power per hour for 
two cases of the above problem? Compare these results with those of Prob. 2, 
>c. 13. 

NOTE: B = .416. 

Prob. 3. What must be the cut-off in a 10- and 15 X20-in. compressed air engine 
running on 100 Ibs. per square inch gage initial pressure and atmospheric back pres- 
sure, to give complete expansion, and what will be the horse-power per 100 ft. per 
minute piston speed, s being 1.4? 

Prob. 4. It is desired to run the following engine at its most economical load. 
What will this load be and how much steam will be needed per hour? 

Cylinders 20- and 32x18 ins., speed 150 R.P.M., steam pressure 100 per square 
inch gage, atmospheric exhaust, dry saturated steam. 

Prob. 5. Should the load increase 50 per cent in Prob. 4, how would the cut-off 
change and what would be th effect on the amount of steam used? 

Prob. 6. What would be gain in power and the economy of the engine of Prob. 4 
were superheated steam used, for which s = 1.3? 

Prob. 7. In a 14- and 20x24-in. engine will the work be equally divided 
between the cylinders for the following conditions? If not, what per cent will be done 
in each? Steam pressure 100 Ibs. per square inch absolute, back pressure 10 Ibs. per 
square inch absolute, s = 1.2, cut-off = 5. 

Prob. 8. What would be the work and steam used by the above engine if there 
were complete expansion and equal distribution? 



306 



ENGINEERING THERMODYNAMICS 



15. Compound Engine with Infinite Receiver. Logarithmic Law. With 
Clearance and Compression, Cycle XI. General Relations between Pressures, 
Dimensions and Work. In terms of point pressures and volumes, Fig. 92, the 




INDICATOR CARDS OF EQUAL BASE 
AND HEIGHT FOR CASE OF INCOM- 
PLETE EXPANSION AND COMPRESSION. 



INCOMPLETE EXPANSION ANCT COMPRESSION. V 




- (- rec-.pr ) R 



(in.pr. 



*4 rel.pr 



bk.pr.) 



( rel.pr.) 




INDICATOR CARDS OF EQUAL BASE 
AND HEIGHT FOR CASE OF OVER 
EXPANSION AND COMPRESSION. 



OVER EXPANSION AND COMPRESSION. 



FIG. 92. Work of Expansive Fluid in Compound Engines with Infinite Receiver, with Clearance, 
Cycle X Logarithmic, and Cycle IX Exponential Expansion, and Compression. 

work of the two cylinders may be written down at once as if each were inde- 
pendent of the other, the connection between them being fixed first by making 
the back pressure of the high equal to the initial pressure of the low, or to 



WORK OF PISTON ENGINES 



307 



he receiver pressure, and second by making the volume admitted to the low 
qual to that discharged from the high reduced to the same pressure. This 
ist condition may be introduced in either of two ways, 



(a) EM = NH, 

(b) [(PV) on H.P. expansion line -(P V) on H.P. comp. line], 

= [(PF) on L.P. expansion line (PV) on L.P. comp. line]. 

Without introducing the last relation 

|^^^ (424) 

^)-PzF;(^ . (425) 



+P, t V h -P l V l -P V +P l V l -PjV J +P k V k . 
The second condition is 



or 



Substituting 



10g e ~ 

Vb 



log, ~PcVe 10g e ^~P*V t log, ^ 
Vh * f 



(426) 



+2(P b V b +P k V k )-P a V a -P g V g -P d V d -PjV J 

This expression, Eq. (427) contains, however, the receiver pressure which is 
ited to the release pressure by 



(rec.prO =P^Pe = P^P. = fi= 
introducing this 

W = 



, log e ~ P * 






(428) 



308 ENGINEERING THEEMODYNAMICS 

Introducing the usual symbols in Eqs. (427) and (428) and in additic 
the following: 

Z = cut-off as fraction of stroke, so that Z H D H is the displacement volun 

up to cut-off. 
c = clearance volume divided by displacement, so that C H D H is the clearan< 

volume and (Z H +c H )D H is the volume in the high-pressure cylindi 

at cut-off. 
X = that fraction of the stroke during which compression is taking place i 

that (X H +Cn)D H is the volume in the high-pressure cylinder whe 

compression begins. 

Applying the general symbols to Eq. (427), 

(in.pr.) (Z H +c H )D H log* (J~^) 
+ (Tec.pr.)(Z L +c L )D L log e 



W = 144 



- (rec.pr.) (X H +c ff )D H log* 



- (bk.pr.) ( 

+2(in.pr.) (Z H +c a )D H - (m.pr.)c H D H - (rec.pr.) 
+2(bk.pr.) (X L +c L )D L - (rec.pr.)c L Z> - (bk.pr.) (1 +c L }D L 



. (4! 



This expression gives the work in terms of initial, receiver and back pressur 
the valve periods, cut-off and compression, the clearances and cylinder c 
placements. 

Substitution of the symbols in Eq. (428) will give another equivalent expr 
sion in terms of the same quantities except that low-pressure cylinder reles 
pressure will take the place of receiver pressure. This is 



/ , log* 

, /l+c L \ /X H +D H \ 

- (rel.pr.)/, ,. (X H +c H )D H \og e I - 

\/L+CZ,/ \ C H ) 

- (bk.pr.) (X L +c L )D L log, 



+2(in.pr.)(Zjsf +c H )D H - (in.pr.)c^D^ - (pel.pr.)z.( J + L - ) 

\ZL +CL I 



WORK OF PISTON ENGINES 309 

'' It is sometimes more convenient to involve the cylinder ratio and low- 
pssure displacement than the two displacements as involved in Eq. (430) 
id the ratios of expansion instead of cut-offs. This may be done by 

iJ T 

V c 



d it should be noted here that the ratio of expansion in each cylinder is no 
iger the reciprocal of its cut-off, as was the case when clearance was zero, nor 
the whole ratio of expansion equal to the product of the two separate ones 
cause the low-pressure cylinder expansion line is not a continuation of that 
the high. Making these substitutions for cylinder and expansion ratios, 
[. (430) becomes, 

log e Rn + (rel.pr.)z,(l +C L ) log e R L 



44D/, 

-^(l +C H ) 



+2(bk.pr.) (X L +C L ) - (relpr.) L R L c L - (bk.pr. 

It is interesting to note that this reduces to Eq. 304 of Section 9, by making 
ance and compression zero. 

From any of the expressions for work, but more particularly (430) and 
52), the usual expressions for (m.e.p.) referred to low-pressure cylinder, work 
3r cubic foot supplied, and consumption per hour per I.H.P. can be found, 
it as these are long they are not set down, but merely indicated as follows: 

W 
(m.e.p. ref. to L.P.) = . .. - - . . (433) 

W 

fork per cu.ft. supplied = 



u.ft. sup. per hr. per I.H.P. 

13,750 , rec. 

~ - 



13,750 -. , Y , f ) 

~ (XL+CL) rec.pr . n.pr. 



310 ENGINEERING THERMODYNAMICS 

As the receiver pressure is related to the initial and back pressures and t< 
the relation between the amount taken out of the receiver to that put in, whicl 
is a function of the compression as well as the cut-off and cylinder ratio, it i 
expressed only by a complicated function which may be derived from th( 
equivalence of volumes in the high and low, reduced to equal pressure. 

PV-P e V e = P h V h ^P t V t , or P ft 
Therefore, 



Tvftv 

Introducing symbols 

c.pr.) = (in.pr.) (ZL+CL )^+^^C^D^ + (bk '^ (Z^c L )D L +(X H +c H )D a 
Hence 

4- (bk nr ^ (^L+c L )R c ,. 

-h^DK.pr.;, \v__i_fY i r^ ^ 4t5b 



This Eq. (436) gives the receiver pressure in terms of initial and back pressures 
the two clearances and compressions, the cylinder ratio and the cut-off in eac 
cylinder. 

Proceeding in a similar way, the release pressures can be found in terms 
initial data, 

17 
p P c 

t C * &T7- J 

v b 



or 

(a) 



(43 

(6) ' 



And 




or 

(Xt+c L ) 

l.pr.), = (m.pr.)|-^^^^ |+(bk.pr.) 




(43! 



WOKK OF PISTON ENGINES 



311 



These three pressures all reduce to those of Eqs. (308), (309), (310), Section 
9, when clearance and compression are zero. 

Equal work in both cylinders is, of course, possible, but it may be secured 
by an almost infinite variety of combinations of clearance, compression and 
cut-off in the two cylinders for various ratios of expansion; it is, therefore, not 
worth while setting down the equation of condition to be satisfied, but reference 
may be had to Eqs. (424) and (425), which must be made equal to each other, the 
result of which must be combined with the equation of cylinder relations. 



\ 



\ 





INDICATOR CARDS OF EQUAL 
BASE AND HEIGHT 



FIG. 93. Special Case of Cycles XI and XII Complete Expansion and Compression in both 
Cylinders, of Compound Engine with Clearance and Infinite Receiver. 

There are certain special cases of this cycle for which equations expressing 
important relations are simpler, and they are for that reason worth investigat- 
ing. Those that will be examined are 

(a) Complete expansion and compression in both cylinders, Fig. 93. 

(&" Complete expansion in both cylinders with no compression, any clearance, 
Fig. 94. 

(c) Any amount of expansion and compression but equal in both cylinders, 
equal clearance percentages and a cylinder ratio equal to the square root of 
the ratio of initial to back pressures, Fig. 95. 

Case (a) When both expansion and compression are complete in both cylin- 
ders, Fig. 93, 

.pr.)^*log. , .... (439) 



A, .... (440) 



312 ENGINEERING THERMODYNAMICS 

but 

Z#D//(in.pr) =Z L D L (rec.pr.) 
and 



(in.pr .)\ /(rec.pr.)\ /(in.pr.) (rec.pr.)\ _ (in.prO 

p7)/ H - logc l(bk^)/~ 10gc l(r^rO (bk.pr.)/ ge (bk.pr.)' 



hence 

'"irriV ' - : -' (441)' 



(442) 



Work per eu.ft. supplied = 144(in.pr.) log, - ....... (443) 



rr 

Consumption, cu.ft. per hr. per I.H.P^^ l__- 5. . . . (444) 

(m.e.p. ref. to L.P.) R c 



Equality of work in high- and low-pressure cylinders is obtained by making 

/ (in.pr. )\ = /(rec-prOx /(in^j_)\ * 
Urec.pr.)/ \(bk.pr.)/ V(bk.pr.)/ ' 



or 

(rec.pr.) = [(in.pr .)(bk.pr.)]* ....... (445) 

It is desirable to know what clearances and displacements will permit of 
equal work and complete expansion and compression. 



hence 



or calling 

(in.pr. ) 



/(Jn-prOx = ^CH_ = /(in.pr. )*\ 
\(rec.pr.)/ Z H +c a \(bk.pr.) /' 
/(rec.pr.)\ = l+c L = /(in.pr. )\ * 
\(bk.pr.)/ Z L +c L \(bk.pr.)/ : 



Equatuig discharge of high and intake of low-pressure cylinders, 

or = R C =R. 



WORK OF PISTON ENGINES 
Inserting in this the values just found for Z H and Z L , 



D , 



313 



(447) 



rtiich is the required relation between cylinder sizes, clearances and ratio of 
>ressures, which, together with cut-offs given in Eq. (446), will give equal work 
nd complete expansion and compression. The compression in the high- 
ressure cylinder is such that 



nd for L.P. cylinder. 

P 



(448) 



L A 



\ 



\ 



H\ 




G H 



INDICATOR CARDS OF EQUAL 
BASE AND HEIGHT 



V 

lop. 94. Special Case of Cycles XI and XII. Complete Expansion and Zero Compression 
in both Cylinders of Compounds Engine with Clearance and Infinite Receiver. 

I Case (b) With complete expansion and no compression, both cylinders, any 
jearance, Fig. 94, 

1 (449) 



T^-'-T) -c [(rec.pr.)-(bk.pr.)]l (450) 
^DK.pr. ;/ j 



T-- 
^DK.pr. 

8th the added requirement_that the high-pressure discharge volume, EC = low 
essure admission volume FH, or 



(451) 



314 ENGINEERING THERMODYNAMICS 

and 



...... (452) 

hence 

[(bk.pr. )] = Zz,+Cz, 

L(rec.pr.)J l+c L ' 
which substituted in Eq. (451) and rearranging gives 

R c = - 



Eq. (453) indicates that for this special case of complete expansion and no com- 
pression the cylinder ratio required to give this case, is determined entirely bj 
the L.P. cut-off and clearance. If the cylinder ratio and clearance are fixed, th(j 
required cut-off in the L.P. cylinder can be found by solving Eq. (453) for Z 2 



Z L =^f^-c L , (454 

tic 



and from Eq. (452), 

(rec.pr.) = (bk.pr.) 



D 
ti 



Cut-off in the high-pressure cylinder is determined by clearance, initia 
pressure and receiver, pressure, which in turn depends on low-pressure cut-oi 
and clearance Eq.(452), or may be reduced to cylinder ratio and low-pressur 
clearance by Eq.(454), as follows 

V c= I+CH = /( in.pr. )\ = /(in.pr. )\Z L +c L 
V* Z H +CH \(rec.pr.)/ \ (bk.pr.)/ l+c L ' 
hence 

,, (l+c s )(l+c L ) 
AH= p (7 - x -- CH- 

Rp(^L-rc L ) 
Eliminate Z L by Eq. (454), 

Re , 

" cir - (45( 



Since the high- and low-pressure cut-offs are functions of cylinder and clearam 
dimensions, and of Rp, the rato 'of initial and back pressures, the work of higl 
and low-pressure cylinders may be expressed entirely in terms of these quantities 



(45! 



WOKK OF PISTON ENGINES 
Hence, total work by addition is 



315 



-c L (R c -l)\. . . (459) 



Expressions might be easily written for mean effective pressure referred 
to the L.P. cylinder, work per cubic foot fluid supplied, and consumption, but 
will be omitted for brevity. It is important to note, however, the volume of 
fluid used per cycle is not AB, but is LB, Fig. 94, and is, 



(Sup.Vol.) = D H \ (Z H +c H ) - C H ^ ^1 = D H [ (Z H +c H ) - cj^ I . . (460) 
L (m.pr.; j L Mr A 

(m.e.p.ref.toL.P.)= ni |^ c ....... (461) 

w 

(Work per cu.ft. supplied) = ( gup y ol y .... (462) 

13 750 f 

Consumption cu.ft per hr. per I.H.P. = / m r ^ f to L P ) &* 



Equality of work, secured by equating Eqs. (457) and (458) gives 

c-c.^e-l). - (464) 



This equation may be satisfied in an infinite number of ways. One case 
worth noting is that of equal clearances, when it is evident that if 



= CL, and = #c, or 



the Eq. (464) is satisfied. This last condition is the same as that which satisfied 
Case (a) with complete compression. 
Case (c), Fig. 95, assumes that 



and 



316 



ENGINEERING THERMODYNAMICS 



and corresponds to the first special case considered in Section 9, which lead 
in the no-clearance case to equality of high- and low-pressure work. 




FIG. 95. Special Case of Cycles XI and XII, Equal Per Cent Clearance in Each Cylinder of 
Compound Engine with Infinite Receiver and Cylinder Ratio Equal to the Square Root 
of Initial Divided by Back Pressure. 

The assumptions already made are sufficient to determine the receiver pres- 
sure. By Eq. (436) 



(rec.pr.) = (in.pr.) 



Z+c 



n. 



+ (bk.pr.) 



[(in.pr.) (bk.pr.)]* 



= [(in.pr.) (bk.pr.)]*. . . . ....... . . . . . 

The work of the high-pressure cylinder may now be evaluated. 

^=144D // (m. P r.){z[n-lo g e (^)1~ 
{ L w~rC/j 

-144D^[(in.pr.) (bk.pr .)]* 



. (465) 



. (466) 



WORK OF PISTON ENGINES 
The low-pressure cylinder work may be similarly stated, 



but 



~D*l=^ 



317 



(467) 



- 144D (bk.pr.) I (X+c) log e ( ~^) +1 -X \ , 

\ c / 



and similarly, 



.) =D*[(in.prO(bk.pr.)] x . 



With these substitutions the value of low-pressure work,W L ,lq. (467), becomes 
equal to high pressure work, Eq. (466), hence the total work 



(468) 



Example 1. Method of calculating Diagrams, Fig. 92. 
Assumed data: 



P q =P a =Pb = 120 Ibs. per square inch abs. 

p n =p g =p e =p d =p h = 50 Ibs. per square inch abs. 

P t =Pj= 10 Ibs. per square inch abs. 



(V h -V n )=(Vm-Ve). 



F=y/=.12cu.ft. 
F 6 = .4cu.ft. 
V c = V d = .Scuft. 
V g = Fi = .16cu.ft. 
F <= V= 2cu.ft. 



V k = Acu.it. 



The above may be expressed in initial pressure, etc., and in terms of cut-off, etc., 
but as the relation of the lettered points to these terms is shown on the diagram values 
for cut-off, etc., they will not be given here, as they may readily be found from values of 
the lettered points. 



To locate point C: 
To locate point F: 
To locate point Q: 



=60 Ibs. per sq.in. 



V f .12 

PeF e= ^0 = 

* Q -r-k -* r\s\ 



per sq.in 



318 ENGINEERING THERMODYNAMICS 

To locate point L: 



-- T7 1C =25 Ibs. persq.m. 

Vi .lo 

To locate point N: 

F n = -= = . 08 cu.ft. 
To locate point 77: 

(7 A _F B ) = (7 OT _F e ), or 7 / , = F m + 7 n -F e = .96+.08-.2 = .84 cu.ft., 
since 

PmV m = P b V b , = V m -~ = .96 CU.ft. 

To locate point 7: 






Example 2. Find (a) the horse-power, (6) steam used per hour, and (c) receiver 
and release pressures for a 12- and 18x24-in. engine with infinite receiver, 6 per 
cent clearance in high-pressure cylinder, and 4 per cent clearance in low-pressure cylinder, 
when initial pressure is 150 Ibs. per square inch absolute > back pressure 10 Ibs. per 
square inch absolute, speed 125 R.P.M., cut-off in high-pressure cylinder is f, low- 
pressure cut-off is such as to give complete H.P. expansion, and compression is 15 per 
cent in high and complete in low. 

(a) For complete high-pressure expansion the receiver pressure must be equal to the 
high-pressure release, and to maintain the receiver pressure constant the low-pressure 
cylinder must take as i-aich steam per stroke as the high-pressure discharges. With 
initial pressure and cut-off as given, the release pre:sure for the high-pressure cylinder 
may be found from the relation (in pr.)(c#+Z#) =(re\.pT.) a (c H -\-D H ) or 150x(.56) 
= (rel.pr.)//(1.04), or (rel.pr.)H = 79.3 Ibs. Since there is 15 per cent compression in 
high-pressure cylinder there is exhausted each stroke 85 per cent of its volume. Also 
since compression in low-pressure cylinder is complete, the low-pressure clearance is 
full of steam at the receiver pressure at the beginning of the stroke. Hence the 
low-pressure displacement up to cut-off must equal .S5D H or L.P. cut-off = .85D H , 
divided by cylinder ratio, or .85 -J- 2.25 = .378. As compression is complete, the 
per cent compression may be found from the relation Cz,x(rec.pr.) =(c L +X L )(bk.pT ), 
or .04X79.3 = (.04+Xz,)10, or Xz, = .28. 

From Eq. (432), (m.e.p.) referred to low-pressure cylinder is obtained by dividing 
by 144 D L , and on substituting the above values it becomes, 



150(.1+.06)( |--) log, 2+30(1 +.04) log, 2.64-30(.15 + .06) (|~)log e (^ 



-10(.28+.04)lo ge 



"264 
-30X^-(1+.06)+2X10(.28+.04)-30X2.64X.04-10(1+.04)=60.5 Ibs. per. sq.in., 

hence 

I.H.P.=235. 



WORK OF PISTON ENGINES 
(6) From Eq. (435) by substituting the above values 

13 750 T / 

u.ft. steam per hour per horse-power = '- ( (. 

o0.5 I \ 



319 



38 +.04) -(.28 +.04) -- -- 
loO 



-1=45.5, 
O J 



pounds per hour will be 3550. 

(c) Release pressure for high-pressure cylinder has been shown to be 79.3 Ibs. 
may be checked by Eq. (437), as follows: 



79.3 Ibs. 



Receiver pressure has already been shown to be equal to this quantity and may 
checked by Eq. (436) 



(rec.pr.)// 



150 X (.5 +.06) 



_ IPX (.28 +.04)2.25 

(.378 + .04)2.25 + (.28+.04) + (.378+.04)2.25 + (.28+.04) 



= 79.3 Ibs. 



Low-pressure release pressure is found from Eq. (438) to be 

.28 + .04 

i _i_ r\A. 9 \xO OF; 
(rel.pr.)/, 



1 + 



(.15 + .06)2.64 



_ (1 +.04)2.25 _ 



+ 10 



1+.04 



.06)2.64 



1+ (1+. 04)2.25 _ 



= 30 Ibs. 



Prob. 1. What will be the horse-power and steam used by the following engine 
or the data as given? 

Engine 20 and 28x36 ins., running at 100 R.P.M., clearance 5 per cent in high 
>ressure, 3 per cent in low. From cards H.P. cut-off = .3, L.P. = .4, H.P. compression, 
1, L.P., .2. Gages show (in.pr.) to be 150 Ibs., (r c.pr.) 60 Ibs., (bk.pr.) 26 ins. 
Ig. (barometer = 30 ins.). 

Prob. 2. What must be ihe cut-offs and the cylinder ratio of an engine to give 
qual work and complete expansion and ompression for 200 Ibs. per square inch 
.bsolute initial pressure and atmosphe: ic exhaust, if clea ance is 5 per cent in the 
igh and 3 per cent in the low-pressure cylinder? What will the horse-power for an 
Bgine with a low-pressure cylinder 24x36 ins., running at 100 R.P.M. for this case? 

Prob. 3. Should there be no compression, how would the results of Prob. 2 be 
|tered? 

Prob. 4. What will be the total steam used by engines of Probs. 2 and 3? 

Prob. 5. For an 11- and 19x24-in. engine with 5 per cent clearance in each 
ylinder, cut-off in each cylinder, and 20 per cent compression in each cylinder, what 
rill be the horse-power and the steam consumption when the speed is 125 R.P.M., the 
nitial pressure 150 Ibs. per square inch gage, and back pressure at atmosphere? 

; 16. Compound Engine with Infinite Receiver. Exponential Law, with 
Clearance and Compression, Cycle XII. General Relation between Pressures, 
dimensions and Work. Referring to Fig. 92, of the preceding section, which 



320 ENGINEERING THERMODYNAMICS 

will represent this cycle by a slight change of slope of the expansion and cor 
pression lines, the high-pressure work may be expressed in terms of dimensior 
ratios and pressures. Since this must contain receiver pressure as a factc 
and since that is not an item of original data, it is convenient first to sta 
receiver pressure in terms of fundamental data: 



But 



V?^ and Vn 



rec.pr./ \rec.pr 

Whence, 



whence 



c.pr./ \rec.pr. 




or in terms of dimensions and pressures, 

i 
(rec.pr.) = (in.pr.)l - R (z 

The high-pressure work may be stated as follows: 




C H 



WOEK OF PISTON ENGINES 



321 



. . (472) 




(473) 



The expression for total work need not be written here, as it is simply the sum 
>f Eqs. (470) and (472) or of (471) and (473), the former containing receiver pres- 
sure and the latter containing only dimensions, initial and back pressures and 
both, the exponent of the. expansion, s, 

The volume of high-pressure fluid supplied per cycle is QB, Fig. 92, which 
may be expressed either in terms of high pressure or of low pressure points, 
thus; 



(Sup.Vol.) 



(474) 



The following quantities will be indicated, and may be evaluated by sub- 
stitution from the preceding: 



W 



W 






W 



Work per cu.ft. fluid supplied = ^ y o \ ) 

sumption cu.ft. per hr. per I.H.P. 

13,750 



(Sup.Vol.) 
(m.e.p.ref.toL.P.) D L 



Equal division of work between high- and low-pressure cylinders requires 
[that Eqs. (470) and (472), or (471) and (473) be made equal. The latter will 



322 ENGINEERING THERMODYNAMICS 

give an expression showing the required relation between dimensions and initial 
and final pressures, cut-off and compression in high- and low-pressure cylinders. 
In this expression there are so many variables that an infinite number of com- 
binations may be made to give equality of work. 

It is desirable to examine the results of assuming special conditions such as 
those of the previous section, the most important of which is that of complete 
expansion and compression in both cylinders, which is represented by Fig. 93. 

- 

.. . . (478) 



81 

' ], . , . (479) 

but 

i_ 

in - r - 



hence 






/rec.pr.\ /jn^pr^y [ 1 _/bk : Pi\ s 1 1 
\ in.pr. / \rec.pr. / [ \rec.pr ./ J J 

[m.prOD^^J^/rec^XT 1 
t \in.pr. / 

S-l 8-1 S -l 

rec.pr. \ s _ / recjjr.X * /bk.pr.X a 
, in.pr./ \in.pr7/ \rec.prT/ J 

s-l 

(480 

The receiver pressure may be found as follows. In Fig. 93, EC = GH: 



Equating 



WORK OF PISTON ENGINES 



323 



When this is solved for receiver pressure it results in an equation of the 
econd degree, which is somewhat cumbersome, and will not be stated here. 
]q. (481) is, however, used later to find Re. 

If work is to be equally distributed between high- and low-pressure-cylinders, 
rom Eqs. (478) and (479), 



S-l 8-1 8-1 S-l 

_ /rec.prA _ /rec.prA * _ /rec.prA /bk.pr^N * 
\ in.pr. / ~ \ in.pr. / \ in.pr. / \rec^prV ' 

S-l S-l 

rec.prA . =1 /bk.pr.\ . 
in.pr. / \in.pr. / ' 






1.0 



25 



50 

tr i f 

Values o 



75 



100 



}. 96. Curves to Show Receiver Pressure to Give Equal Work Distribution when Expansion 
and Compression are Complete in both Cylinders of the Compound Engine with Infinite 
Receiver, with Clearance when Expansion and Compression are not Logarithmic. 



ice, for equal division of work, 



(rec.pr.) = (in.pr.) 



/bk-pr.\Vl^i 
\in.pr./ 



(482) 



lich, if satisfied, will give equality of work in the two cylinders, for this case 
perfect compression and expansion. 

In Fig. 96, is given a set of curves for use in determining the value of the 
>io of (rec.pr.) to (in.pr.) as expressed by Eq. (482). 



324 ENGINEERING THERMODYNAMICS 

When (rec.pr.) has been found by Eq. (482) it is possible by means of (481) 
and the clearances to find Re- The events of the stroke must have the follow- 
ing values to maintain complete and perfect compression and expansion. 



: (483) 



(484) 



(486) 



Example. Find (a) the horse-power, (b) compressed air used per hour, and (c) receiver 
and relea.e pressures for a 12- and 18x24-in. engine with infinite receiver, 6 per centi 
clearance in the high-pressure cylinder, and 4 per cent in the low-pressure cylinder, when 
initial pressure is 150 Ibs. per square inch absolute, back pressure 10 Ibs. per square 
inch absolute, speed 125 R.P.M., cut-off in high-pressure cylinder , low-pressurei 
cut-off such as to give complete expansion in high-pressure cylinder, compression inj 
high-pressure cylinder 15 per cent, and complete in low. Expansion such that s = 1.4.; 

(6) As in example of Seztion 15, receiver pressure equal high-pressure release 
pressure, and low-pressure volume at cut-off must equal volume of steam exhaustec 
from high pressure. Release pressure may be found from relation (m.pY.J(cH+Z H ) 
= (rel.pr.) (c#+Z)//) s . or 150(.06+.5) 14 = (rel.pr.) (.06+1) 1 ' 4 , or(rel.pr.) =60 Ibs. A 
in the previous example, the low-pressure cut-off is .38, and the low-pressure compressior 
may be found from the relation c/, s (rec.pr.) = (cz,+Xz,) s (bk.pr.),or(.04) 14 x60 
(.04+Xz,) 1 ' 4 (10), or X L = .09. 

From the sum of equations (471) and (473) divided by U4D H Rc, and with th 
proper values substituted the following expression for (m.e.p.) results; 

I/ 10 \ .71 \ i. 
.5 + .06 + 2.25(.15 + .06)(-) 

2.25(.38+.04)+.15 + .06 

.71 1 1. 



^ (^)'N+--*M : 

.4 L\ .06 / ( 2.25(.38 + .i 

f.38+.04r /.38 + .04V 4 ] 

L'-ni+wj . -< 



.04 
~ar - 1 + 1 -- 09 =55 Ibs. persq.in. 



hence the horse-power is 214. 



WORK OF PISTON ENGINES 325 

From Eq. (477) with proper values substitute, 



Cu.ft.perI.H.P. hr.^pxTu 
oo (_ 

total steam per hour will be 

50X214 = 10700 cu.ft. 

(c) Release for the high-pressure cylinder has already been given as 60 Ibs. and the 
ceiver pressure the same. The latter quantity may be checked by equation (469) 
id will be found to be the same. The low-pressure release pressure may be found 
jm the relation (rec.pr.)(Z i +cz / ) 1 - 4 = (rel.pr.) z/ (l+c z ,) 1 - 4 , which on proper substitution 
ves 

.38+.04\fi 



(rel.pr.)z,=60( r ) =27 Ibs. per sq. inch 
\ 1 .U4 / 

| Prob. 1. What will be the horse-power and steam used per hour by an 18- and 

X30-in. engine with 5 per cent clearance in each cylinder and with infinite receiver 
inning on 100 Ibs. per square inch gage initial pressure, and 5 Ibs. per square inch 
:>solute back pressure, when the speed is 100 R.P.M. and the cut-off in high-pressure 
linder is \ and in low T 4 W ? 

NOTE: s = 1.3 and 8 = .2. 

Prob. 2. What must be the receiver pressure for equal work distribution when the 
iitial pressure has the following values for a fixed back pressure of 10 Ibs. per 
tuare inch absolute? 200, 175, 150, 125, 100, and 75 Ibs. per square inch gage? 

Prob. 3. For the case of 150 Ibs. per square inch gage initial pressure and 14 Ibs. 
If square inch absolute back pressure, what will be the required high-pressure cylinder 
peforan air engine with a low-pressure cylinder 18x24 ins., to give equality of work, 
jjarance in both cylinders being 5 per cent? 

Prob. 4. What will be the horse-power and air consumption of the above engine 
len running at a speed of 150 R.P.M. , and under the conditions of perfect expansion 
d compression? 

, 17. Compound Engine with finite Receiver. Logarithmic Law, with 
I'earance and Compression, Cycle XIII. General Relations between Pressures, 
mensions, and Work when H.P. Exhaust and L.P. Admission are Inde- 
i-ndent. As this cycle, Fig. 97, is made up of expansion and compression lines 
jferred to the different origins together with constant pressure, and constant 
lume lines, the work for high- and low-pressure cylinders and for the cycle can 
j set down at once. These should be combined, however, with the relation 
S'ted for the case of infinite clearance which might be termed the condition 
T a steady state 

[(PV) on H.P. expansion line]-[(PF) on H.P compression line] 
= [(PV) on L.P. expansion line PV on L.P. compression line,] 

P h V b -PeV e = P fl V fl -P t V t (487) 



326 



ENGINEERING THERMODYNAMICS 



INCOMPLETE EXPANSION ANC 





(in.pr.) L 



(c.o.pr.)L 
I (rel.pr.) L 






l(rel.pr.) u |< 



OVER EXPANSION AND COMPRESSION 



FIG. 97. Work of Expansive Fluid in Compound Engine with Finite Receiver and w 
Clearance. Cycle XIII, Logarithmic Cycle XIV, Exponential Expansion, and Co 
pression. 



WORK OF PISTON ENGINES 327 

Besides this there is a relation between H.P. exhaust and L.P. admission 
pressures, corresponding to the. equality that existed for the infinite receiver, 
ithat may be set down as follows: 

P m V m = PV, and P m (V m +0)=P d (V d +0)', 
.'. P m V m = P b V b = P d (V d +0)-P m O, and P m = P h ', 

p b v b +p h o 



*~ v d +o 

Also 



( } 



, and P n (V n +0)=P g (V g +0); 
-P n O, and P n = P e , 



V a +0 






These two expressions for the pressure at D and at G are not available in 
leir present form, since they involve two unknown pressures those at H and 
, but two other equations of relation can be set down from which four equa- 

jtions, the four unknown pressures P e , P d) P g and Pn, can be found. These 

mother equations are 

Pt(V d +0)=P e (V e +0), or p r P.(Mg), . . (490) 
;and 

P,(V a +0)=P*(V h +0), or p,=p h (. . . (491) 

Equating (488) to (490), 



*- = '> or 

;and 

p _P b V h +P h O 

Ve + 

Equating (489) to (491) 

P t V*+PeO T>(V+0\ PV4-pn-p(v 

-(y^oT = A, or P t V t +P e O-P h (V h 

and 



Therefore D _ 

JLe 



Ve+0 

P t ,V b O+P h 2 =P h (V h +0)(Ve+0)-P t V t (Ve+0); 



V h Vc+V e O+V h O ' 



328 



Therefore 



ENGINEERING THERMODYNAMICS 

n PV b O+P k V k (V c +0)^ . (fl) 



Substitution will give 

Pe 



(V e +0)(V h +0)-0 2 
"^7 6 0+P t F t (F e +0)l [^0] 

* T7/T7 L^^y e Q [V + 0\ 



(6) 
(c) 



_ rpy(7+Q)+p,7 t o] [ 

d ~ -0 2 J I 



. (492) 



It will be found that the use of these pressures is equivalent to the applica- 
tion of the equation of condition given in Eq. (487), for substitution of them 
reduces to an identity, therefore the work of the two cylinders can be set down 
by inspection in terms of point pressures and volumes and the above pressures 
substituted. The result will be the work in terms of the pressures and cylinder 
dimensions. 



= P 6 7 6 (l+loge ^) -PeVe-PeV e loge ^-PeF e O ge ^ 

Therefore 



!Fjr-PF(l+log, ^) - 



J \ ' O, 



(4-: ' 



-P.V, log, 

Therefore 



log, -P*F t log, - 



t rp & 7 ft o+p t y,(Fe+Q) 



(494) 



WORK OF PISTON ENGINES 



329 



Adding W H and W L 



W = 



L(^ + 0)(F + 0)- 



V( 



(495) 



F/ 

While this Eq. (495) for the cyclic work is in terms of initial data, it is not 
of very much value by reason of its complex form. To show more clearly 
i that only primary terms are included in it, the substitution of the usual symbols 
will be made. 



TF=144X 



(bk.pr.) (X L +c L )D L log e ^-^ 
CL 

(in.pr.)c//D# - (bk.pr.) (1 X L )D L 

(w^pr^ZH +CH)DH[(Z L +c L }D L +0] + (bk.pr.) (X L +c^)Dz,0 
+0}[(Z L + CL )D L +0]-0* 

tf+ca)i>H+Ql 



[(X a +C H )DH +0] 



(in.pr.) (Z H +c H )D H + (bk.pr. 



I 



(Z L +CL)DL[(X H +CH)DH 

\(Z L +c L )D L +0~] 
c L D L +0 



(in.pr.) (Z H +c H }D H + (bk.pr.) [(X L 



(Z L +c L )D L 



H +0] + (X H +c H )D H 



X 



X 



X 



. (496) 



Such equations as this are almost, if not quite, useless in the solution of 
problems requiring numerical answers in engine design, or in estimation of engine 
performance, and this fact justifies the conclusion that in cases of finite receivers 
graphic methods are to be used rather than the analytic for all design work. When 
^estimates of power of a given engine are needed, this graphic work is itself seldom 
Justifiable, as results of sufficient accuracy for all practical engine operation 
problems can be obtained by using the formulas derived for infinite receiver when 
'reasonably, large and zero receivers when small and the pistons move together. 



330 



ENGINEERING THERMODYNAMICS 



It might also be possible to derive an expression for work with an equivalent 
constant-receiver pressure, that would give the same total work and approxi- 
mately the same work division as for this case, but this case so seldom arises that 
it is omitted here. 

Inspection of the work equations makes it clear that any attempt to find 
equations of condition for equal division of work for the general case must be 
hopeless. It is, however, worth while to do this for one special case, that of 
complete expansion and compression in both stages, yielding the diagram Fig. 
98. This is of value in drawing general conclusions on the influence of receiver 
size by comparing with the similar case for the infinite receiver. 

By referring to Fig. 98, it will be seen by inspection that cylinder sizes, 
clearances and events of the stroke must have particular relative values in order 




FIG. 98. Special Case of Cycles XIII and XIV, Complete Expansion and Compression in 
both Cylinders of Compound Engine with Clearance and Finite Receiver. 

to give the condition assumed, i.e., complete expansion and compression. It! 
is, therefore, desirable to state the expressions for work in terms which may 
be regarded as fundamental. For this purpose are chosen, initial pressure 
(in.pr.), back pressure (bk.pr.); high-pressure displacement, D H ] cylinder 
ratio, RC] high-pressure clearance, C H ; and ratio of receiver volume to high- 
pressure displacement, y. Call 



bk.pr. 



= n, P . 



It will be convenient first to find values of maximum receiver pressure 
(rec.pr.)i, and minimum (rec.pr.) 2 ; high-pressure cut-off Z H , and compressioi 
X H ; low-pressure clearance CL, cut-off Z L , and compression, XL, in terms o 
these quantities. Nearly all of these are dependent upon the value of CL am 
it will, therefore, be evaluated first. 



WORK OF PISTON ENGINES 331 

From the points C and /, Fig. 98, 

From A and E, 

(rec.pr.)i = (in.pr.)-^- , (498) 

RcCi! 

and from E and C, 



(rec.pr.) 



(rec.pr.) 2 R c c L +y ' 
Dividing Eq. (498) by Eq. (497) and equating to Eq. (499), 

C H ) = 1+Cn+y 






Multiplying out and arranging with respect to C L , the relation to be 
fulfilled in order that complete expansion and compression may be possible is, 



= Q. (500) 
This is equivalent to 



n = 0, ....... (501) 

and the value of CL is 



(502) 



It is much simpler in numerical calculation to evaluate I, m, and n and 
insert their values in Eq. (502) than to make substitutions in Eq. (500), which 
would make a very cumbersome formula. (rec.pr.)2 and (rec.pr.)i may 
now be evaluated from Eqs. (497) and (498) by use of the now known value of C L . 

High-pressure cut-off ZH may be found from the relation of points B and 
J, Fig. 98, 



tip 
Low-pressure cut-off, Zi L from, 



or 

ZH= ^(I+ CL )- CH (503) 

or 

Z L J^L- CL (504) 

tic 

High-pressure compression, X H , 

C H (505) 



332 ENGINEERING THERMODYNAMICS 

Low-pressure compression, X L , by the use of points A and K t 

Rc(X L +c L )=R P c H , 
or 

X L = ^C H -C L ......... (506) 

He 

If C L is regarded as being part of the original data, though it is related to 
R c , R P , C H and y as indicated in Eq. (488), the expressions for high- and low-pres- 
sure work and may be stated as follows: 



= 144(in.p r O 



. (508) 
Adding these two equations gives the total work of the cycle as follows: 

W= 144(in.pr.)Z>*ftc(l+c t ) f i [l+log, f? ^1 +~ log, 

(ttp\_ -n:c(l+C/JJ /tp 



1+cu+y } g . H } 

" ' 10ge 



log* 



This, however, may be greatly simplified, 

logc 
and 

10gc C H ' "* e R c c L 
Hence 



R P . 

(599) 



WORK OF PISTON ENGINES 



333 



From this may be obtained mean effective pressure referred to the low- 
pressure cylinder, work per cubic foot supplied, and consumption per hour 
per indicated horse-power, all leading to the same results as were found for the 
case of complete expansion and compression with infinite receiver (Section 15,) 
and will not be repeated here. 

To find the conditions of equal division of work between cylinders, equate 
Eqs. (507) and (508). 



lich may be simplified to the form, 



l+cn+y , , fi*p(l+c g ). RPCH \ } #W J _ 

T&I^ R^' 

(510) 

_.iis equation reduces to Eq. (376) of Section 11, when CH and CL are put 
equal to zero. In its present form, however, Eq. (510) it is not capable of 
solution, and it again becomes apparent that for such cases the graphical solution 
of the problem is most satisfactory. 

Example 1. Method of calculating Diagram, Fig. 97. 
Assumed data: 



P a =P b = 120 Ibs. per square inch abs. 
p m =p h = 30 Ibs. per square inch abs. 
PJ= 10 Ibs. per square inch abs. 



To locate point C: 



To locate point M : 






P h V h 120 X. 4 
30 



Vj=V t = 2 cu.ft. 
V d = Vt= -8 cu.ft. 
V g = Vi= .24 cu.ft. 

y e = .2 cu.ft. 

V a = Vf= .12 CU.ft. 

y =i.2 cu.ft. 

F 6 = .4 cu.ft, 



60 Ibs. per sq.in. 



= 1.6 cu.ft. 



To locate point D: 

Pa(V d +0)=P m (V m +0}, or 



-= 42 Ibs. per sq.in. 



334 ENGINEERING THERMODYNAMICS 

To locate point E: 

2 8 
P e = (V e +0)=P m (V m +0), or Pe =^^3X30 =60 Ibs. per sq.in. 



To locate point F: 
To locate point L: 
To locate point N: 



p = 

K/ 



since P n =Pe 

To locate pcint G: 

P a (V +0) =Pn(V n +0] or, P^O^;;; =55.5 Ibs. per sq.in. 

To locate point H: 



or 

V, . .24X55.5+55.5X1.2-30X1.2 _ ^ ^ ft 

To locate point /: 



Example 2. Find the horse-power of a 12- and 18x24-in. engine, running at 125 
R.P.M., with a receiver volume twice as large as the low-pressure cylinder, 6 per cent 
clearance in the high-pressure cylinder, 4 per cent in the low, when the initial pressure 
is 150 Ibs. per square inch absolute, back pressure 10 Ibs. per square inch absolute, high- 
pressure cut-off \ t low-pressure f , high-pressure compression 10 per cent, low 3 per cent. 
From Eq. (496) divided by 144Z>z,, and with the values as given above, the (m.e.p.) 
is equal to following expression: 



150 X .56[.79 X2.25 +4.5] +10(.34)2.25 X4.5 1 .16 1.06X2.25+4J 

[.16+4.5][.79x2.25+4.5]-(4.5) 2 2.25 ge .06 X .16 X 



150 X .56 X4.5 + 10 X .34 X2.25[.16 +2.25] ".79x2.25+4.5 .79X2.25+4J 



] 
J'' 



.79 X2.25[.16 +2.25] + (.16 X2.25) [-04 X2.25 +4.5J ' 4 X2 ' 25 loge .04 X2.25 +4.; 
150X.56X4.5+10(.34X2.25)[.16+2.25][V 1.041 

f .79x2.25[.16+4.5] + (.16x2.25) [ ^ ^~J =4 ' 2 lbS " ^ ^^ 

hence the horse-power will be 191. 



WORK OF PISTON ENGINES 335 

Prob. 1. Find the work done in the high-pressure cylinder and in the low-pressure 
cylinder of the following engine under the conditions given. 

Engine 14 and 30x28 ins., 100 R.P.M., 5 per cent clearance in each cylinder, 
high-pressure cut-off i 3 o, low-pressure cut-off T 4 o, high-pressure compression TO, low- 
pressure compression TO, initial pressure 100 Ibs. per square inch gage, back pressure 
5 Ibs. per square inch absolute, and receiver volume 3 times the high-pressure 
displacement . Logarithmic expansion. 

Prob. 2. The following data are available: initial pressure 200 Ibs. per square inch 
absolute, back pressure 10 Ibs. per square inch absolute, engine 10x15x22 ins., with 
5 per cent clearance in the high- and low-pressure cylinders, speed 100 R.P.M. What will 
3e the cut-offs, and compression percentages to give complete expansion and compression. 
Logarithmic expansion? 

Prob. 3. What will be the work done by the above engine working under these 
conditions? 

Prob. 4. What must be the low-pressure clearance, cut-offs, and compression 
[percentages, to give complete expansion and compression for a similar engine work- 
ing under the same conditions as those of Prob. 2, but equipped with a receiver twice 
!is large as the high-pressure cylinder? 

18. Compound Engine with Finite Receiver, Exponential Law, with 
Clearance and Compression, Cycle XIV. General Relations between 
[Pressures, Dimensions, and Work when H.P. Exhaust and L.P. Admission 
lire Independent. It cannot be expected that the treatment of this cycle by 
formulas will give satisfactory results, since even with the logarithmic expansion 
|w, Cycle XIII gave formulas of unmanageable form. For the computation of 
work done during the cycle, however, and for the purpose of checking pressures and 
work determined by graphical means, it is desirable to have set down the relations 

dimensional proportions, initial and final pressures, and valve adjustments, to 

receiver pressures, release pressures and work of the individual cylinders. 

The conditions of a steady state, explained previously, require that (Fig. 97) 



firhich is the same as to say, that the quantity of fluid passing per cycle in the 
ligh-pressure cylinder must equal that passing in the low. Expressed in terms 
if dimensions, 




i 

- n 7? ( . 7 }\ ( cu ^"ff prQ^I _ 

( in.pr.) 

>r, rearranging, and using R P = rr-r^ N 

(DK.pr.; 



tm^r^H ~rnc \i<Li~r.t*- LI "77 i \^n i ^*-HJ \ / 

R P - ,_ (m.pr 



j. 

(in.pr.) J i 

4_7? t r 7 x[" (cut-off pr.)z,] x. 12 v 
+#cfe+^- ( . npr) J , .(512) 



336 ENGINEERING THERMODYNAMICS 

an equation which contains two unknown pressures (rec.pr.)i and (cut off pr.) L 
To evaluate either, another equation must be found: 

i 

/p k \ 

' n = Ptl T- 

where P n = P c , so that 



Hence 

v + 



or 

- (, , Y J(bk.pr.) I.- 

fo+^ 

(cut-off pr.) t = (rec-pr.)! 



_ _ a 

[ y (rec.pr.) * i + R c (c L +X L ) (bk.pr.) 1 ' 

L y+flcfo+Zi) J, V ' 



which constitutes a second equation between (cut-off pr.)z, and (rec.pr.)i, which 
used with Eq. (512) makes it possible to solve for the unknown. By substitu 
tion in Eq. (512) and rearranging, 



ir 
(rec.pr.)i- bk.pr.) |rcLciL 

This expression is of great assistance even in the graphical constructio 
of the diagram, as otherwise, with all events known a long process of trial an 
error must be gone through with. It should also be noted that when s = l th 
expression does not become indeterminate and can, therefore, be used to solv 
for maximum receiver pressure for Cycle XIII, as well as Cycle XIV. 

Cut-off pressure of the low-pressure cylinder, which is same as the pressure L 
H or at M, Fig. 97, is now found most easily by inserting the value found b 
Eq. (515) for (rec.pr.)i in Eq. (514). 

Enough information has been gathered now to set down the expressions fc 
work. 



WORK OF PISTON ENGINES 337 



. . (517) 



Addition of these two Eqs. (516) and (517) gives an expression for the total 
work TF, and equating them gives conditions which must be fulfilled to give 
equality of work in the high- and low-pressure cylinders. Since these equations 
I so obtained cannot be simplified or put into more useful form, there is no object in 
inserting them here, but if needed for any purpose they may be easily written. 
In finding the conditions of equal work, the volumes of (rec.pr.)i and (cut-off pr.)^ 
must be inserted from Eqs. (514) and (515) in (516) and (517), in order to have 
terms in the two equations consist of fundamental data. This, however, increases 
greatly the complication of the formula. 

After finding the total work of the cycle, the mean effective pressure referred 
to the low pressure is obtained by dividing by 144X-D.L. 

To assist in finding the work per cubic foot supplied and consumption, 
and the cubic feet or pounds per hour per I.H.P. it is important to know the 
volume of fluid supplied per cycle, 

i 
(rec 



ec.pr.)A "] 
E^F.)/ J- 



Example. Find the horse-power of and compressed air steam used by a 12- and 
18x24-in. engine running at 125 R.P.M., with a receiver volume twice as large as the 
low-pressure cylinder, 6 per cent clearance in the high-pressure cylinder, 4 per cent in 
the low, when the initial pressure is 150 Ibs. per square inch absolute, back pressure 
10 Ibs. per square inch absolute, high-pressure cut-off |, low-pressure cut-off f, 
high-pressure compression 10 per cent, low-pressure compression 30 per cent, and 
expansion and compression follow the law PV lA =c. 

From Eq. (515) (rec.pr.)i is found to be as follows when values for this problem 
are substituted: 



^ r)l , lo 

-16 [4.5 +2.25 X .79] +4.5 X2.25 X .79 



and by using this in Eq. (514) 

.71 .71 



. 

/4.5X81.7 +2.25X.34X10 \ __ ., . , ., 

(cut-off pr.)z, = (- 4. 5+2 .25^79" ~ ) =53 lbs * 8q>m - absolute ' 



338 ENGINEERING THERMODYNAMICS 

It is now possible by use of Eqs. (516) and (517) by addition and division by 
144D* to obtain (m.e.p.). Substituting the values found above and carrymg out 
the process just mentioned. 




-10x2.25X.7 =51.5 Ibs. sq.in. 



hence the horse-power will be 200. 

By means of Eq. (518) the supply volume may be found. This gives upon 
substituting of the proper values: 



(Sup.Vol.) =L 

L 

13,750 Sup.Vol. 
Cubic feet per hour per I.H.P. = ^ ^X ^ , 

_ 13,750 .46 
= 51.5 X 2.25" 

hence the total volume of air per hour will be 

54.5X200 = 10900 cu.ft. 

Prob. 1. What will be the receiver pressure and L.P. cut-off pressure for < 
cross-compound compressed air engine with 5 per cent clearance in each cylinder, run 
ning on 100 Ibs. per square inch gage initial pressure and atmospheric exhaust, whei 
the high-pressure cut-off is i, low-pressure f, high-pressure compression 15 per cent 
low 25 per cent, and s = 1.4. Receiver volume is twice the high-pressure cylinde 
volume. 

Prob. 2. Find the superheated steam per hour necessary to supply a 14- and 21 X28 
in. engine with 5 per cent clearance in each cylinder and a receiver twice 1 the siz 
of the high-pressure cylinder when the initial pressure is 125 Ibs. per square inch gage 
back pressure 7 Ibs. per square inch absolute, speed 100 R.P.M., high-pressur 
cut-off , low-pressure T 4 o, high-pressure compression 15 per cent, low pressure 40 pe 
cent and s = 1.3. 

NOTE: 8 = .3. 

Prob. 3. If the high-pressure cut-off is changed to I without change of any othe 
factor in the engine of Prob. 2, how will the horse-power, total steam per houi 
and steam per horse-power per hour be affected? If it is changed to f ? 

Prob. 4. A boiler capable of supplying 5000 Ibs, of steam per hour at rated load fur 
nishes steam for a 12- and 18 x24-in. engine with 5 per cent clearance in each cylinder an* 
running at 125 R.P.M. The receiver is three times as large as the high-pressure cylindei 



WORK OF PISTON ENGINES 



339 



the initial pressure 150 Ibs. per square inch gage, back pressure 5 Ibs per square inch 
absolute, the low-pressure cut-off fixed at \ and low-pressure compression fixed at 
30 per cent. At what per cent of its capacity will boiler be working for these follow- 
ing cases, when =1.2 for all and 20% of the steam condenses during admission ? 
(a) high-pressure cut-off I, high-pressure compression 80 ] er cent, 
(6) high-pressure cut-off \, high-pressure compression 20 per cent, 
(c) high-pressure cut-off |, high-pressure compression 10 per cent. 
NOTE: = .33. 




INDICATOR CARDS OF EQUAL BASE 
AND HEIGHT 




P 



A' A 



\ 



(in.pr.) 






--(rel.pr.) H 
(in. pr.) L 



QH 



Pr.)i 



X L D 




alz 



E" 



K 



I 




l+CH+KcC|.1 

Rc-1 
(bk.pr.)- 



IM 



-r, ri+CH+RcCul 

" DL L Rc-i J 



FIG. 99. Work of Expansion in Compound Engine without Receiver and with Clearance. 
Cycle XV, Logarithmic Expansion; Cycle XVI, Exponential, High-pressure Exhaust 
and Low-pressure Admission Coincident. 

19. Compound Engine without Receiver. Logarithmic Law, with Clear- 
iance and Compression, Cycle XV. General Relations between Pressures, 
'Dimensions, and Work when H.P. Exhaust and L.P. Admission are Coinci- 
; jdent. The graphical construction for this cycle has been described to some 
extent in connection with the first description of the cycle, given in Section 8, of 
.this chapter, and is represented here by Fig. 99 in more detail. 

To show that the expansion from D to E is the same as if volumes were 
bieasured from the axis ML, consider a point Y on DE. If the hypothesis 
is correct 

(519) 



340 ENGINEERING THERMODYNAMICS 

The true volume when the piston is at the end D of the stroke, i: 
D H (l+c a +RcCL), and at Y, the true volume is 

D a (l+c a +Rcc L )-D H y+D L y, 

where y is the fraction of the return stroke that has been completed in bo 
cylinders when the point Y has been reached. Then 

P d Dn(l + CH +RCCL) = P V D H (\ +c H +R c c L ) +P v D H (Rc+l}y. 



Dividing though by (R c 1), 

- 1 . (520) 

J 



__This equation may be observed to be similar in form to Eq. (519). More 
over, the last term within the bracket, D H y, is equal to the corresponding tern 
Y'K, in Eq. (519), hence, 

,- 01 
' ' ' ' (521 



Similarly, the distance QM, or equivalent volume at D' is 

. . ..." (522 



The following quantities will be evaluated preparatory to writing the expra 
sions for work: 



v 



...... (522 

, (bk.pr.) 



SC 
(in.pr.)i=P' = (rel.pr.)ff 



- . 1 

1/7 . T Hctip -- -T ^ - ; -- 

Z "+ C ^ 1 y (m.pr.) Z tf +c H J 
l 



1-fc// _ __ . .. . 4 

- N (bk.pr.) 

\-K c (&L-rCL)-(-' 

(52. 



(cutoff p,),= (in. P ,), ^ 



WORK OF PISTON ENGINES 
The ratio of expansion from E' to G is equal to 
(cut-off pr.)z, I+CL 



341 



(rel.pr.)z, 



(527) 



Hence 



V H = 144D,,(m.pr.) { (Z H +c H ) \l +log e 



- 
ZB+CH\ 



Z H +c H +Rc(X L +cz,) 



(m.pr. ) 



Rc-l 

(bk.pr.)" 



(CH+XH) loge CH 



CH 



(528) 



\\z 

Jn.pr.) _^ 



x 



1+C. 



CL ) 



. . . (529) 



The total work found by adding W H and W L as given above, leads to the 
following : 



^(in.pr.) j (ZH+CH) Iog 6 ( - 



m.pr. 



-144Dx,(bk.pr.) j 1 - 



loge 



. . . (530) 



342 ENGINEERING THERMODYNAMICS 

This is the general expression for the work of the ^compound engine without 
receiver, with clearance and compression, when high-pressure exhaust and low- 
pressure admission are simultaneous and expansion and compression logarithmic, 
in terms of fundamental data regarding dimensions and valve periods. 

From this the usual expressions for mean effective pressure, work per 
cubic foot supplied, and consumption per hour per I.H.P., may be easily 
written, provided the supply volume is known. This is given by 



(Sup. Vol.) =A?B = D H [(Z a +c a ) - 



. . (531) 

To find the conditions which must be fulfilled to give equal work in the two 
cylinders, equate Eqs. (528) and (529). 




0. (532 



These expressions are perfectly general for this cycle, and expressed ii 
terms of fundamental data, but are so complicated that their use is ver 
limited, as in the case of some of the general expressions previously derivec 
for other cycles. 

As in other cycles, it is desirable to investigate a special case, that o 
complete expansion and compression in both cylinders, Fig. 100. First it i 
necessary to determine what are regarded as fundamental data in this case, an 
then to evaluate secondary quantities in terms of these quantities. The followin 
items are assumed to be known: (in.pr.), (bk.pr.), which is equal to (rel.pr.)j 
R c , C H , and C L , and D H , which are dimensions, and it is known that tb 
pressure at the end of compression in L.P. is equal to (rel.pr.)^. 

Referring to the diagram, displacements, clearances, and the axis for 
common expansion, ML, can all be laid out, and the location of the poinl 
A and G determined. 

The points E and E' are at the end of the common expansion within tl 
two cylinders, and at beginning of high-pressure compression and of separal 
low-pressure expansion, hence p e = pe. 



WOEK OF PISTON ENGINES 
From the points A and E: 

fr-frH-Ozfe 

From the points G and E' ', 

/V=pe = (bk.pr.)- 



343 




M V 



|IG. 100. Special Case of Cycles XV and XVI. Complete Expansion and Compression in 
both Cylinders of Compound Engine without Receiver and with Clearance High- 
Pressure Exhaust and Low-Pressure Admission Coincident 

and equating, 



Whence 



[where 



(533) 



Substituting the value of X H in either of the expressions for p e , which is 
he low-pressure cut-off pressure, 



(cut-off 



(534) 



344 ENGINEERING THERMODYNAMICS 

It may be noted here that the cylinder ratio does not enter into this, but 
only clearances and pressures. In the no-clearance case, it may be remembered 
that the point E or E' was not present, as it coincided with G. 

Next, to find the high-pressure release pressure, p dj by means of points E 
and Z), and their relation to the axis ML, Fig. 100. 




.RC(I~^-CL) -\~RpCn /COK\ 

-(bk.pr.) - ,-> - (535) 



Knowing the release pressure of the high-pressure cylinder, it is possible tc 
find the high-pressure cut-off and compression necessary to give the required 
performance. 

,(rel.pr.) g 



p I_L^ -LP n 
(in.pr.) HP L i+CH-\-ticCL J 

(rel.pr.) g \ 

/i i ^ -- CL = CL\ 
(bk.pr.) L 



(rel.pr.) g \Rc(l+c L )+RpCH J 

= -- = ~ -- 1 

J 



The work of the two cylinders is as follows: 
w 



^cci,] \Rc(i+c L )+Rpc H ~\ j_ , 

-l JL 1+CH+RcCL jRp ^ 



Re 

' ) 



R P (R 



+fipcir], rg c (i+te) 

-l) J * L l+c^+ 



I+C L + CH 



WORK OF PISTON ENGINES 345 

These expressions, when added and simplified, give the following for total 
work per cycle, 

( r r> /-i i _ \ i r> _ ~~i 'i 

. . (540) 



in which of course D H R C may be used instead of D L and ^ ?P-^ instead of 

K P 

(bk.pr.) and then 

p, . ... (541) 



Z H having the value of Eq. (536). 

Equality of work the in high- and low-pressure cylinders results, if W u 
Eq. (538) equals W L , Eq. (539), or if 



, or 

all of which lead to equivalent expressions. Simplification of these expres- 
sions, however, does not lead to any direct solution, and hence the equations 
will not be given here. 

Example. Find (a) the horse-power and (6) steam used per hour for a 12- and 18 x24-in. 
tandem compound engine with no receiver, 6 per cent clearance in the high-pressure 
cylinder, and 4 per cent in the low, when the initial pressure is 150 Ibs. per square inch 
absolute, back pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., high- 
pressure cut-off , high-pressure compression 15 per cent and low-pressure compression 
is complete. 

(a) Since the low-pressure compression is complete, the pressure at end of compression 
must be equal to the release pressure of the high. This latter quantity may be found 
*rom the relation (in. 



/ r * Y 

(rel.pr.)^ = 150 -r-= 79.3 Ibs. per sq.in. absolute. 
1 .uo 

Low-pressure compression may be found from the relation (rel.pr.)tf(cz,) = (bk.pr.) 
(CL+XL), or X L = .28. (m.e.p.) may be found from Eq. (540) divided by 144Dz,, which 
on substitution gives 



.5+.06+2.25(.28 + .04)- 



+ . ...oo. ______ o 2.25U.04-.15) log e 



1 +.06+2.25 X.04 + 1.75+.85J * 1+.04-.15 



(.06 + .15) loge ~ -10 1 1.28 + (.28+.04) loge ^~^ \ =69.7. Ibs. sq. in. 

.Ob J ( 

the horse-power will be 271. 



346 ENGINEERING THERMODYNAMICS 

(6) Since the consumption in cubic feet per hour per horse-power is equal to 

'_13,750_ Sup.Vol. 
(m.e.p.) DL 

and supply volume is given by Eq. (531), this becomes 



69.7 "2.251 \1.06 + .09 + 1.25X.85 

hence the consumption per hour will be 

44 X271 X.332 =4000 pounds. 






Prob. 1. A Vauclain compound locomotive has cylinders 18 and 30x42 ins., with 5 
per cent clearance in each and runs on a boiler pressure of 175 Ibs. per square inch gage 
and atmospheric exhaust. The steam pressure may be varied as may also the cut-off to a 
limited degree. For a speed of 200 R.P.M., a cut-off f and 10 per cent compression 
in each cylinder, find how the horse-power will vary with the initial pressures of 175, 
150, 125, and 100 Ibs. gage. 

Prob. 2. When the cut-off is reduced to in the above engine compression 
increases in the high-pressure cylinder to 20 per cent- For the case of 175 Ibs. gage 
initial pressure find the change in horse-power. 

Prob. 3. Find the steam used by the engine per hour for the first case oi 
Prob. 1 and for Prob. 2. 

Prob. 4. It is desired to run a 12- and 18x24-in. no-receiver engine with 5 per cenl 
clearance in each cylinder, under the best possible hypothetical economy conditions for ar 
initial pressure of 200 Ibs. per square inch absolute and atmospheric exhaust. To 
give what cut-off and compression must the valves be set and what horse-powe] 
will result for 100 R.P.M.? 

20. Compound Engine Without Receiver. Exponential Law, with Clear- 
ance and Compression, Cycle XVI. General Relations between Pressures 
Dimensions, and Work, when H.P. Exhaust and L.P. Admission are Coin- 
cident. Again referring to Fig. 99, it may be observed that reasoning 
similar to that in Section 19 but using the exponential law, would show thai 
the same formulas and graphical constructions will serve to locate th( 
axes of the diagram, hence, as before, 



-, (542 

and 



= DS ' rr^T 7 ^. . (543 

Kc~~ 1 

Release pressure in the high-pressure cylinder is 

)' (544 



WORK OF PISTON ENGINES 



347 



Immediately after release the pressure is equalized in the high-pressure 
cylinder and the low-pressure clearance. The pressure after equalization, 
termed (in.pr.) L , is found by the relation of the volume at S and that at D', 
Fig. 99, measured from the axis KT. 




which, by means of Eq. (544) becomes 



The expansion of the fluid goes on as it passes from the high-pressure 
cylinder to the greater volume in the low-pressure, as indicated by D'E' 
>and DE, and when the communicating valve closes, the pressure has become 




(cut-off pr.) L = (in.pr. 



which, by means of Eq. (545) reduces to 

i 



After cut-off in the low pressure, expansion goes on in that cylinder alone 
to the end of the stroke, when release occurs at a pressure 



(rel.pr.)z, = (cut-off pr 
r by substitution from equation (546), 



\1* (547) 
)\ ' 



nr ), - ft>k r.r ) 

k - pr0 L l+c H +RcC L +(l-X H )(R c -l) 

In terms of these quantities the work of the high- and low-pressure 
cylinders can be written out as follows: 



frir = 144D* 



s-1 

(bk.pr.) 



'r / I+CH+RCCL -V" 1 

I \l+c H +Rcc L +(l-X H )(R c -l) / 

+^y- 1 ill /-ox 

7~y 



348 ENGINEERING THERMODYNAMICS 

and 

(bk.pr.) 



RC-I r 

x\i- 

\l+c H +R c c L +(l-X)(Rc- 



s-1 

I 



. (bk.pr.) , J R*P(CH+ZH)+RC(CL+X L } ] s |~ /l + c L -X H \'-^ 

L - .1 J^ O- ^^ // ) I ~" \ ~"\" I I J. """" I I 

:.pr.) . (549) 



These are general expressions for work of high- and low-pressure cylinders 
for this cycle, and from them may be obtained the total work of the cycle, 
mean effective pressure referred to the low-pressure cylinder, and by equating 
them may be obtained the relation which must exist between dimensions, 
events, and pressures to give equal division of work. It would, however, 
be of no advantage to state these in full here, as they can be obtained from 
the above when needed. 

The supply volume, cubic feet per cycle, is represented by A 'B, Fig. 99, 
and its value is found by referring to points B and E as follows: 

(Sup.Vol.) =D a \(c a +Z a ) - (c a +X a ) ((cut-off P r.)A!-| , 

\ (m.pr.) / J 

-n L 4.7 (CH+X H ) ( R> P(CH+Z H )+RC(CL+X L ) \1 . . 
~T- ( l+ c a +Rcc L +(l-X H )(R c -T)) J ' (55 

Work per cubic foot supplied is found from Eqs. (548), (549), and (550). 



Work per cu.ft. supplied = .. . . . . . >. . . (551) 



Consumption, cubic feet per hour per I.H.P., is found from mean effective 
pressure referred to L.P. cyl. and supply volume as follows: 

Consumption, cu.ft. per hr. per I.H.P. 

13,750 (Sup.VoL) I 

(m.e.p. ref. to L.P.) D L 

This will give pounds consumption by introducing the factor of density. | 
Further than this, it will be found more practicable to use graphica 
methods instead of computations with this cycle. 



WORK OF PISTON ENGINES 349 

Example. Find (a) the horse-power and (6) consumption of a 12- and 18x24-in. 
no-receiver engine having 6 per cent clearance in the high pressure cylinder and 4 per 
cent in the low when the initial pressure is 150 Ibs. per square inch absolute, back 
pressure 10 Ibs. per square inch ab.iolute, speed 125 R.P.M., high-pressure cut-off , 
high-pressure compression 15 per cent, and low-pressure compression is complete. 

(a) The per cent of low-pressure compression may be found as in the Example of 
Section 19, using the value of s in this case of 1.4. Then 



or 

(rel.pr.)#=61.5 Ibs. sq. inch absolute, 
and 



or 



!. 

From the sum of Eqs. (548) and (549) divided by 144Z>z, and with proper 
alues substituted, (m.e.p.) =48.5 Ibs.; hence the horse-power is 189. 

13 750 

(b) From Eqs. (550) the value for (Sup. Vol.), which when multiplied by - , 

m.e.p. 

and divided by D L , gives cubic feet air per hour per I.H.P. 



!LZ?y J-F 
48.5 2.25[ 



15- 7 (.56)+2.25(.15) _] =. 63 cu . ft . per 



(15)' 7 1 +.06+2.25 X.04 + (l -.15)(2.25-1) hourper I.H.P. 



Prob. 1. If the locomotive of Prob. 1, Section 19, should be equipped with superheater 
o that the steam expanded in such a way that s = 1.3, what would be the effect upon 
tie horsu-power for conditions of that problem and on the cylinder event pressures? 

Prob. 2. A 30- and 42x54-in.no receiver steam pumping engine runs at 30 R.P.M. 
nd has 3 per cent clearance in the high-pressure cylinder and 2 per cent in low. There 
is no compression in either cylinder. Initial pres ure is 120 Ibs. per square inch gage, 
md back pressure 28 ins. of mercury (barometer reading 30 ins.). The steam is such 
bat the expansion exponent is 1.25. What will be the horse-power of, and the steam 
ised by the engine when the cut-off in the high is ^? 

Prob. 3. By how much would the power change if the cut-off were shortened to f 
and then to i, and what would be the effect of these changes on the economy? 

21. Triple-Expansion Engine with Infinite Receiver. Logarithmic Law. 
So Clearance, Cycle XVII. General Relations between Pressures, Dimen- 
sions and Work. Fig. 101 represents the cycle of the triple-expansion engine 
with infinite receiver, no clearance, showing one case of incomplete expansion 
p. all cylinders, and another where overexpansion takes place in all cylinders. 



350 



ENGINEERING THERMODYNAMICS 



The reasoning which follows applies equally well to either case, and to an; 
combination of under or overexpansion in the respective cylinders. 

It is desired to express the work of the respective cylinders and th 
total work in terms of dimensions, initial and back pressures, and the cut-off 
of the respective cylinders. To do this, it is convenient first to express th 



B E F 

H.P.V I. PA 
\ D \ 



(1st rec.pr-) INDICATOR CARDS OF EQUAL 




INCOMPLETE EXPANSION 



V 

CYCLE XVII 



OVER EXPANSION 



FIG. 101. Work of Expansive Fluid in Triple-Expansion Engine with Infinite Receiver i 
Zero Clearance. Cycle XVII, Logarithmic Expansion. 

first receiver pressure (1st rec.pr.) and second receiver pressure (2d rec.pr.) 
terms of these quantities. The subscript / refers to the intermediate cylind 



or 



and 



/i j_ \ / \ ^H-L) ff 

(1st rec.pr.) = (in.pr.) " , 

rFi 



(55 



or 



(2d rec.pr.) = (in.pr.)^^. 



(55 



WOEK OF PISTON ENGINES 
Work of high-pressure cylinder is 



351 



144(in.pr.)A, 



Work of intermediate cylinder is 



(555) 



^j-f~4 (556) 



Work of low-pressure cylinder is 



. . . (557) 



The total work by addition is 



D 



*-&\-uw**** 



, . (558) 



Mean effective pressure referred to the low-pressure cylinder is found by 
[dividing W by 144D/,, and is therefore 

i.e.p. ref. to L.P.) 



352 ENGINEERING THERMODYNAMICS 

Work done per cubic foot supplied is equal to W divided by the supp 
volume Z# or Z H D H , 






Work per cu.ft. supplied 

= 144(in.pr.) ( 3 +log c ~jr- -%- ~- - 144(bk.pr.)TF- . (5ft 
LH^I^L AiDi ALL)L} &HL)H 

The volume of fluid supplied per hour per indicated horse -power is 



Consumption, cu.ft. per hr. per I.H.P. 

13,750 Z H D 



(m.e.p. ref. to L.P.) D L 



(56 



The weight of fluid used per hour hour per indicated horse-power is 
course found by multiplying this volume Eq. (561) by the density of the flu 
used. 

The conditions which will provide for equal division of work between tl 
three cylinders may be expressed in the following ways: 



which is equivalent to, first: 
or 

'Oge - 1/A 

hence 



Similarly from 



These two equations, (562) and (563), show the necessary relations betf | 

7 7 7 ( Dl \ ( DL \ /(in.pr.)\ 

Ztf, Z 7 , Zz,, (7^-), {-FT , and 7^-^41, 
\D*I WJT/ \(bk.pr.)/' 

in order that work shall be equally divided. Since there are six independ 
quantities entering (as above) and only two equations, there must be } 
of these quantities fixed by conditions of the problem, in order that the ot 



WORK OF PISTON ENGINES 



353 



may be found. For instance, if the cylinder ratios, the pressure ratio, 
and one cut-off are known, the other two cut-offs may be found, though the 
solution is difficult. 

Again, if cut-offs are equal, and the ratio of initial to back pressure is known, 
it is possible to find the cylinder ratios. This forms a special case which is of 
sufficient importance to require investigation. 

If Zn = Zi = Z L , Eq. (562) becomes 



D 



and Eq. (563) reduces to 



but from Eq. (564), 



and therefore 



and 



D I D L (m.pr.) 
(bk.pr.)' 



D 2 DH DH \D H ' 



D 



D 



which, along with the condition assumed that 



Z H = Zj = 



(564) 



(565) 



(566) 



(567) 



lonstitute one set of conditions that will make work equal in the three cylinders 
frhis is not an uncommon method of design, since by merely maintaining 
equal cut-offs, the work division may be kept equal. 

The work done in any one cylinder under these conditions Eqs. (566) 

|ind (567) is then 



nd the total work 



- (568) 



which Z represents the cut-off in each cylinder, all being equal. 



354 



ENGINEERING THERMODYNAMICS 



A special case of the triple-expansion engine with infinite receiver and n< 
clearance which demands attention is that of complete expansion in al 
cylinders, represented by Fig. 102. Here 

D t _ (bk.pr.) 



_> /y=: (2d rec.pr.) 
' Z), (1st rec.pr.)' 




FIG. 102. Special Case of Cycle XVIII Complete Expansion in Triple-expansion Engii 
with Infinite Receiver, Zero Clearance, Logarithmic Expansion. 

and 

7 _ (1st rec.pr.) _ /bk.pr A D L 
(in.pr.) -U-prjA/ 



hence the receiver pressures are as follows: 

(1st rec.pr.) = (bk.pr.; 



(57: 



WORK OF PISTON ENGINES 355 



(2drec.pr.)=(bk.pr.)^ ........ (574) 

The work of the respective cylinders, expressed in terms of initial and 
)ack pressures and displacements is then, 



,,, 



Similarly, 

e \J)ff/ } w'w 



and 

Total work, by addition, is 



(57g) 



If for this special case of complete expansion equality of work is to be 
jfctained, then from Eqs. (575), (576), and (577), 

(in.pr.) D H = D I= D L 
(bk.pr.) D L D H DS 

Bich is readily seen to jbe the same result as was obtained when all cut-offs 
lere equalized, Eqs. (564) and (565). This case of complete expansion and 
ual work in all cylinders is a special case of that previously discussed where 
fct-offs are made equal. Hence for this case cut-offs are equal, 



7-7-7 _K_I_L.PT. 

^--- 



Example. A triple-expansion engine 12- and 18- and 27x24-ins., with infinite 
ceiver and no clearance, runs at 125 R. P.M. on an initial pressure of 150 Ibs. 
|r square inch absolute, and a back pressure 10 Ibs. per square inch absolute. 



356 ENGINEERING THERMODYNAMICS 






If the cut-offs in the different cylinders, beginning with the high, are J, |, and I, 

what will be (a) the horse-power, (6) steam consumed per hour, (c) release and 

receiver pressures? 

(a) From Eq. (559) 

-(bk.pr.), 



hence 

'39X2X573X250 



33,000 

(6) From Eq. (561). 

13,750 Z H D H , 

Cubic feet per horse-power per hour = f N 

(m.e.p.) D L 



hence total pounds per hour will be, 

34.9 X.338X.332 =3920. 
(c) From Eq. (553) 

1st (rec.pr.) = (in.pr.)-^-^, 



= 150 X ' =89 Ibs. per sq.in absolute. 



From Eq. (554) 

2d (rec.pr.) =(in.pr. 



= 150 X ' =3.75. per sq.in absolute. 
.0/0 Xo.uo 

High-pressure release pressure may be found from relation (in.pr.) ZnDu 
(rel.pr.)//D# .*. (rel.pr.)#=75 Ibs. Similarly 1st (rec.pr.)Z/D/ = (rel.pr.)/D/, 01 
(rel.pr.)/=33.4. Similarly 2d (rec.pr.)Z L D L = (rel.pr.)iZ>i,, or (rel.pr.)i = 14.8. 

Prob. 1. What would be the horse-power and steam used per hour by a 10- am 
16- and 25x20-in. infinite receiver, no-clearance engine, running at 185 R.P.M 
on an initial pressure of 180 Ibs. per square inch gage and atmospheric exhaust 
Cut-offs .4, .35, and .3. 

Prob. 2. The following data are reported for a test of a triple engine: 

Size 20x33x52x42 ins., speed 93 R.P.M., initial pessure 200 Ibs. per square incl 

gage, back pressure one atmosphere, H.P. cut-off .5, horse-power 1600, steam per hsore 



WORK OF PISTON ENGINES 357 

ower per hour 17 Ibs. Check these results, using cut-offs in other cylinders to give 
pproximately even work distribution. 

Prob. 3. What change in cylinder sizes would have to be made in the above engine 
3 have equal work .with a cut-off of in each cylinder, keeping the high pressure the 
ime size as before? 

Prob. 4. What would be the horse-power of a triple-expansion engine whose low- 
ressure cylinder was 36x3 ins., when running on 150 Ibs. per square inch absolute 
litial pressure and 10 Ibs. per square inch absolute back pressure, with a cut-off in 
ach cylinder of .4 and equal work distribution? Make necessary assumptions. 

Prob. 5. A triple engine 18x24x36x30 ins., running at 100 R. P.M. on an initial 
ressure of 200 Ibs. per square inch absolute and back pressure of 20 Ibs. per 
quare inch absolute, is to be run at such cut-offs as will give complete expansion in 
11 cylinders. What will these be, what receiver pressures will result, what horse- 
;ower can be produced under these conditions, and how much steam will be needed 
'er hour? 

NOTE: 8 for 200 Ibs. = .437. 

22. Multiple-Expansion Engine. General Case. Any Relation between 

pylinder and Receiver. Determination of Pressure Volume-Diagram and 

iVork, by Graphic Methods. It is possible to arrange multiple-expansion 

jngines in an almost infinite variety of ways with respect to the pressure-volume 

Shanges of the fluid that take place in their cylinders and receivers. There 

pay be two or three cylinder compounds of equal or unequal strokes, pistons 

loving together by connection to one piston rod, or separate piston rods 

dth a common cross-head or even with completely independent main parts 

nd cranks at 0, 180, displaced with either one leading, or the pistons may 

ot move together, being connected to separate cranks at any angle apart, 

ad any order of lead. Moreover, there may be receivers of large or small 

ie, and there may be as a consequence almost any relation between H.P. 

.scharge to receiver and low-pressure receipt from it, any amount of fluid 

assing to correspond to engine load demands and consequently any relation 

cut-offs, compressions, and receiver-pressure fluctuations. Triple and 
uadruple engines offer even greater varieties of combination of related factors, 
i that problems of practical value cannot be solved by analytical methods 
anything like the same ease as is possible by graphic means, and in 
ime cases not at all. 

These problems that demand solution are of two classes. 

1. To find the work distribution and total work for cylinders of given 
imensions, clearances, receiver volumes and mechanical connection or move- 
Lent relation, with given initial and back pressures, and given valve gear at 

setting of that valve gear or at a variety of settings. 

2. To find the cylinder relations to give any proportion of the total work 
any cylinder at any given valve setting or any fraction of initial pressure 
any value of release pressure or total number of expansions. 

The essential differences between these two classes of problems is that in the 
t the cylinder dimensions are given, while in the second they are to be found. 



358 ENGINEERING THERMODYNAMICS 

In general, however, the same methods will do for both with merely 
change in the order, and in what follows the dimensions of cylinder 
valve periods, receiver volume, initial and back pressures will be assume 
and the diagrams found. By working to scale these diagrams will give tt 
work by evaluation of their area, by means either of cross-section paper directb 
on which strips can be measured and added, or by the planimeter. Thi 
will high- and low-pressure work be evaluated through the foot-pound equivi 
lent per square inch of diagram, and the total work or the equivalent mea 
effective pressure found by the methods of mean ordinates referred to th 
pressure scale of ordinates. 

In the finding of the pressure-volume diagram point by point there is but on 
common principle to be applied, and that is that for a given mass the produc 
of pressure and volume is to be taken as constant (for nearly all steam prot 
lems, which is the almost sole application of this work) and when two masst 
come together at originally different pressures and mix, the product of the resul 
ing pressure and the new volume, is equal to the sum of the PV products of th 
two parts before mixture. At the beginning of operations in the high-pressur 
cylinder, a known volume of steam is admitted at a given pressure and i1 
pressure and volume are easily traced up to the time, when it communicate 
with the receiver in which the pressure is unknown, and there difficulty 
encountered, but this can be met by working from the other end of tr. 
series of processes. The low-pressure cylinder, having a known compressio 
volume at the back pressure, there will be in it at the time of opening 
receiver a known volume, its clearance, at a known compression pressur 
The resulting receiver pressure will then be that for , the mixture. These 
receiver pressures are not equal ordinarily, but are related by various compre 
sions and expansions, involving high- and low-pressure cylinder parti 
displacements, grouped with receiver volumes in various ways. 

Take for an illustrative example the case of a two-cylinder, single-actin 
cross-compound engine with slide valves, cylinder diameters 12f and 20 in 
with 24 ins. stroke for both. High-pressure clearance is 10 per cent, 
pressure clearance 8 per cent. Receiver volume 4000 cu.ins. High-pressui 
crank following by 90. Find the mean effective pressure for the high- and k 
pressure cylinders, for a cut-off of 50 per cent in the high, and 60 per ce 
in the low, a compression of 10 per cent in the high and 20 per cent in the lo 
initial pressure 105 Ibs. per square inch gage, back pressure 5 Lbs. p 
square inch absolute, expansion according to logarithmic law. 

On a horizontal line SZ, Fig. 103, lay off the distances 

TU = low-pressure cylinder displacement volume in cubic inches to scale. 
U 7= low-pressure cylinder clearance volume in cubic inches to scale. 
VW = receiver volume in cubic inches to scale. 

WX = high-pressure cylinder clearance volume in cubic inches to scale. 
XY = high-pressure cylinder displacement volume in cubic inches to scale 



WORK OF PISTON ENGINES 



359 



Through these points draw verticals produced above and below, T'T", 
V'U", V'V", W'W", X'X", Y'Y". Then will W'W and WZ be PV 
coordinates for the high pressure diagram in the quadrant W'WZ, and V'V 




'IG. 103. Graphical Solution of Compound Engine with Finite Receiver and with Clearance 
Illustrating General Method of Procedure for any Multiple Expansion Engine. 

id FS the PV coordinates for the low pressure diagram in the reversed 
ladrant F'FS. 
Lay off AB to represent the high-pressure admission at a height XA rep- 



360 ENGINEERING THERMODYNAMICS 

resenting absolute initial pressure; lay off LM at a height TL representing 
low-pressure exhaust at a constant absolute_back pressure to the same scale. 

Locate point B at the cut-off point AB = .5QXY on the initial pressure 
line, and drop a vertical BB 2 and draw similar verticals JJ 2 , GG 2 , MM 2 , at 
suitable fractional displacements to represent L.P. cut-off, H.P. and L.P. 
compression volumes resprectively. 

This operation will fix two other points besides the points A and L, B the 
H.P. cut-off at the initial pressure and M the low-pressure compression at the 
back pressure. Through the former draw an expansion line BC and through 
the latter a compression line MN, locating two more points, C and N, at the 
end of the outstroke of the high and instroke of the low. 

At point C the H.P. cylinder steam releases to the receiver of unknown 
pressure, and at N, the L.P. cylinder steam is opened to both the receiver and 
high-pressure cylinder at unknown pressure and volume. 

To properly locate these pressures and volumes from the previously known 
pressures and volumes hi a simple manner, the construction below the line 
SZ is used. 

Lay off on W'W" the high-pressure crank angles 0-360, and to the right 
of each lay off from the clearance line XX " the displacement of the piston at the 
various crank angles for the proper rod to crank ratio, locating /the curve 
A 2 B 2 C 2 E 2 F 2 G 2 H 2 . This is facilitated by Table XIII at the end of the 
chapter, but may be laid out graphically by drawing the crank circle and 
sweeping arcs with the connecting rod as radius. 

Opposite H.P. crank angle 270 locate L.P. crank angle = 360 and 
draw to left of the low-pressure clearance line UU" the crank angle dis- 
placement curve for that piston. 

It will be noted that steam volumes are given in the lower diagram by the 
distances from either of these curves toward the other as far as circum- 
stances call for open valves. Thus H.P. cylinder volumes are distances from 
the H.P. displacement curve to WW", but when H.P. cylinder is in communi- 
cation with receiver, the volume of fluid is the distance from H.P. displacement 
curve to VV", and when H.P. cylinder, receiver and L.P. cylinder are all 
three in communication the volume is given by the distance from H.P. 
displacement curve to L.P. displacement curve. This pair of displacement 
curves located one with respect to the other as called for by the crank angle 
relations, which may be made to correspond to any other angular relation, 
by sliding the low up or down with respect to high-pressure curve, serve 
as an easy means of finding and indicating the volumes of fluid occupying 
any of the spaces that it may fill at any point of either stroke. 

On each curve locate the points corresponding to valve periods by the 
intersection of the curve with verticals to the upper diagram, such as BB 2 . 
These points being located, the whole operation can be easily traced. 

At H.P. cut-off (B) the volume of steam is B 3 B 2 . During H.P. expansion 
(B to C) steam in the high increases in volume from B 3 B 2 to C 3 C 2 . 

During H.P. release (C to D) the volume of steam in the high C 3 C 2 is 



WORK OF PISTON ENGINES 361 

m 

added to the receiver volume C 4 C 3 , making the total volume C 4 C 2 . During 
H.P. exhaust (D to E) the steam volume C 4 C 2 in H.P. and rec. is com- 
pressed to volume I 3 E 2 . 

At L.P. admission (N) in low and (E) in high, the volume I 2 ! 3 is added, 
making the total volume I 2 E 2 in high, low, and receiver. 

During (7 to Q) in low and (F to G) in high the volume I 2 E 2 in high, low 
and receiver, changes volume until it becomes Q 2 G 2 in high, low, and receiver. 

At H.P. compression, G in the high, the steam divides to Q 2 G 3 in low 

and receiver, while G 3 G 2 remains in high and is compressed to H 3 H 2 , at the 

beginning of admission in the high. The former volume Q 2 G 3 , in low and 

receiver, expands to J 2 J 4 , at the moment cut-off occurs in the low, which 

divides the volume into, J 3 J 4 in receiver, which remains at constant volume 

1 till high-pressure release, and the second part, J 2 J 3 in the low, which expands 

[fin that cylinder to K 2 K 3 . 

After low-pressure release the volume in low decreases from K 2 K 3 to 
M 2 M 3 , when the exhaust valve closes and low-pressure compression begins. 

During compression in low, the volume decreases from7kf 2 M 3 to I 2 ! 3 which 
is the volume first spoken of above, which combines with I 3 E 2 , causing the drop 
in the high-pressure diagram from E to F. 

The effects upon pressures, of the various mixings at constant volume 
between high, low, and receiver steam and the intermediate common expan- 
sions and compressions may be set down as follows: 

At C, steam in high, at pressure P CJ mixes with steam in receiver at 
pressure Pj, resulting in high and receiver volume at pressure P d . 

From D to E there is compression in high and receiver resulting in 
pressure P e . 

At E steam in high and receiver at pressure P e mixes with steam in low, 
at P n , locating points I in low and F in high at same pressure. 

From (F to G in high) and (/ to Q in low) there is a common compression- 
expansion in high, low, and receiver, the pressures varying inversely as the total 
volume measured between the two displacement curves. At G in there begins 
compression in high alone to H. 

In the low and receiver from Q to / there is an expansion and consequent 
fall in pressure from P a to Pj. 

After low-pressure cut-off at J the expansion takes place in low-pressure 
cylinder alone, to pressure P*, when release allows pressure to fall (or rise) 
to exhaust pressure Pi. 

When cut-off in low occurs at / the volume J 3 J 4 is separated off in the 
leceiver, where it remains at constant pressure P/ until high-pressure release 
at point C. 

tAt the point M compression in low begins, increasing the pressure in low 
one from P m to P n . 
There are, it appears, plenty of relations between the various inter- 
ediate and common points, but not enough to fix them unless one be first 
established. One way of securing a starting point is to assume a compression 



362 ENGINEERING THERMODYNAMICS 

pressure P g for the beginning of H.P. compression and draw a compression 
line HG through it, produced to some pressure line a/, cutting low-pressure 
compression line at d. Then the H.P. intercept (e-f) must be equal to the 
low-pressure intercept (d-c); this fixes (c) through which a PV = const, line 
intersects the L.P. cut-off volume at /. 

Now knowing by this approximation the pressure at /, the pressure may 
by found at D, E, F, and at G. The pressure now found at G may differ 
considerably from that assumed for the point. If so, a new assumption for 
the pressure at G may be made, based upon the last figure obtained, and 
working around the circuit of pressures, J, D, E, F, and back to G should 
give a result fairly consistent with the assumption. If necessary, a third 
approximation may be made. 

It might be noted that this is much the process that goes on in the receiver 
when the engine is being started, the receiver pressure rising upon each release 
from the high, closer and closer to the limiting pressure that is completely reached 
only after running some time. 

These approximations may be avoided by the following computation, 
representing point pressures by P with subscript and volumes by reference 
to the lower diagram. Pj is the unknown pressure in receiver before high- 
pressure release and after low-pressure cut-off. 

Pressure after mixing at D is then 



(C 4 C 2 ) 
The pressure at F, after mixing is 



(PE 2 ) 

(PE 2 } 
This pressure multiplied by W gives P 8 , and this in turn multiplic 



by 



Writing this in full, 



Solving for P h 



WORK OF PISTON ENGINES 363 

which is in terms of quantities all of which are measurable from the diagram. 
While this formula applies to this particular case only, the manner of obtain- 
ing it is indicative of the process to be followed for other cases. 

When there are three successive cylinders the same constructions can be 
used, the intermediate diagrams taking the position of the low for the com- 
pound case, while the low for the triple may be placed under the high and off-set 
from the intermediate by the volume of the second receiver. In this case 
it is well to repeat the intermediate diagram. Exactly similar constructions 
apply to quadruple expansion with any crank angle relations. 



Prob. 1. By means of graphical construction find the horse-power of a 12- and 18 X24- 
in. single-acting cross-compound engine with 6 per cent clearance in each cylinder, 
if the receiver volume is 5 cu.ft., initial pressure 150 Ibs. per square inch gage, back 
pressure 10 Ibs. per square inch absolute, speed 125 R.P.M., high-pressure compression 
30 per cent, low pressure 20 per cent, high pressure cut-off 50 per cent, low pressure 
40 per cent, high-pressure crank ahead 70, logarithmic expansion, and ratio of red 
to crank 4. 

Prob. 2. Consider the above engine to be a tandem rather than a cross-compound 
; 'and draw the new diagrams for solution. 

Prob. 3. A double-acting, 15- and 22 x24-in. compound i engine has the high- 
pressure crank ahead by 60, and has 5 per cent clearance in the low-pressure cylinder, 
10 per cent in the high, and a receiver 4 times as large as the high-pressure cylinder. 
What will be the horse-power when the speed is 125 R. P.M. /initial pressure 150 Ibs. per 
square inch absolute, back pressure 5 Ibs. per square inch absolute, high-pressure cut- 
off ^, low-pressure |, high-pressure compression 20 per cent, low-pressure 30 per cent, 
and ratio of rod to crank 5. Determine graphically the horse-power in each cylinder. 

Prob. 4. Consider the engine of Prob. 3 to be a tandem compound and repeat 
the solution. 



23. Mean Effective Pressure, Engine Power, and Work Distribution and 
their Variation, with Valve Movement and Initial Pressure. Diagram Dis- 
tortion and Diagram Factors. Mechanical Efficiency. The indicated power 
developed by a steam engine is dependent upon three principal factors piston 
displacement, speed, and mean effective pressure. The first, piston displacement, 
is dimensional in character, and, fixed for a given engine. Speed is limited by 
steam and inertia stresses, with which the present treatment is not concerned, 
or by losses due to fluid friction in steam passages, a subject that will be 
further considered under steam flow. Mean effective pressure is a third factor 
which is to be investigated, most conveniently by the methods laid down in 
the foregoing sections. 

In these formulas for mean effective pressure, it will be observed that 
the terms entering are (a) initial pressure, (6) back pressure, (c) cut-off or 
ratio of expansion, (d) clearance, and (e) compression, for the single-cylinder 
engine. It is desirable to learn in what way the mean effective pressure 
varies upon changing any one of these factors. 



364 ENGINEERING THERMODYNAMICS 

Referring to Section 5, Eq. (262) for logarithmic expansion 

(m.e.p.) = (in. pr.) Z+ (Z+c)log c ^~- (mean forward pressure) 



-(bk.pr.)\l- 



X+(X+c) 



J 



(mean back pressure) 



(582) 



it is seen that the mean effective pressure is the difference between a mean 
forward pressure and a mean back pressure. The former depends on 
initial pressure, cut-off, and clearance, and the latter on back pressure, 











I 








I 












| 




JRWA 




P 




a 


i 


1 




















































M 


FAN 


F( 


3D 


R 


b 


























M 


FJR 




















Al 


F. 


P. 




















1 


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FIG. 104. Curve to Show Variation of Mean Forward and Mean Back Pressure for Logarithmic 
Expansion and Compression in a Single Cylinder Engine with Clearance. 

compression, and clearance. To study the effect of varying these terms it 
is most convenient to draw curves such as are shown in Fig. 104, and examine 
mean forward pressure and mean backward pressure separately. 

Mean forward pressure is seen by inspection to vary in direct proportion 
to initial pressure. Cut-off, when short, gives a low mean forward pressure, 
but it is to be noted that zero cut-off will not give zero mean effective 
pressure so long as there is clearance, due to expansion of steam in the 
clearance space. Increasing the length of cut-off, or period of admission, 
increases mean forward pressure, but not in direct proportion, the (m.f.p.) 
approaching initial pressure as a limit as complete admission is approached. 



WORK OF PISTON ENGINES 



365 



earance has the tendency as it increases, to increase the mean forward 
pressure, though not to a great extent, as indicated by the curve Fig. 104. 

Mean back pressure is usually small as compared to initial pressure, though 
a great loss of power may be caused by an increase of back pressure or com- 
pression. Back pressure enters as a direct factor, hence the straight line through 
the origin in the figure. So long as compression is zero, back pressure and 
mean back pressure are equal. When compression is not zero, there must 
be some clearance, and the ratio of (mean bk.pr.) to (bk.pr.) depends on both 
clearance and compression, being greater for greater compressions and for 
i smaller clearances, 



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FIG. 105. Curves to Show Variation of Mean Effective Pressure for Logarithmic Expansion 
and Compression in a Single Cylinder Engine with Clearance. 

The mean effective pressures obtained by subtracting mean back from 
mean forward pressures in Fig. 104 are shown in curve form in Fig. 105. 

The multiple expansion engine can not be so simply regarded. In a general 
Way each cylinder may be said to be a simple engine, and subject to variations 
if mean effective pressure due to change in its own initial pressure and back 
pressure, clearance, cut-off and compression, which is true. At the same time 
|hese factors are interrelated in a way that does not exist in the simple 
gine. Consider, for instance, the high-pressure cylinder of a compound 
gine with infinite receiver, with clearance. An increase of high-pressure 
mpression tends first to raise the mean back pressure according to the 
ning on simple engine, but at the same time the change has decreased 
e volume of steam passing to receiver. No change having been made 



366 ENGINEERING THERMODYNAMICS 

in the low-pressure cylinder, the volume admitted to it will remain the same 
as before, and the receiver pressure will fall, decreasing mean back pressure 
by a greater amount than compression increased it, and mean forward being 
the same as before, the increase of high-pressure compression has increased 
the mean effective pressure of the high-pressure cylinder. The only effect 
upon the low-pressure cylinder is that resulting from lowering its initial 
pressure, i.e., the receiver pressure. This results in a decrease of low-pressure 
mean effective pressure. Computation will show that the assumed increase 
of high-pressure compression decreases low-pressure work more than it 
increases high-pressure work, or in other words, decreased mean effective 
pressure referred to the low. 

It is impracticable to describe all results of changing each of the 
variables for the multiple-expansion engine. Initial pressure and cut-off 
in the respective stages have, however, a marked influence upon receiver 
pressures and work distribution which should be noted. Power regulation 
is nearly always accomplished by varying initial pressure, i.e., throttling, 
or by changing cut-off in one or more cylinders. 

The effect of decreasing initial pressure is to decrease the pressures on 
the entire expansion line and 'in all no clearance cycles, to decrease absolute 
receiver pressures in direct proportion with the initial pressure. Since back 
pressure remains constant, the result is, for these no-clearance cycles, that 
the mean effective pressures of all but the low-pressure cylinder are decreased 
in direct proportion to the initial pressure, while that of the low-pressure is 
decreased in a greater proportion. The same is true only approximately 
with cycles having clearance and compression. 

The conditions giving equal work division have been treated in connection 
with the individual cycles, it may here be noted in a more general way that 
if high-pressure cut-off is shortened, the supply capacity of that cylinder is 
decreased, while that of the next cylinder remains unchanged. The result 
is that the decreased supply volume of steam will be allowed to expand to 
a lower pressure before it can fill the demand of the next cylinder than itl 
did previously, i.e., the receiver pressure is lowered. Similarly shortening 
cut-off in the second cylinder will tend to increase receiver pressure. To 
maintain constant work division, there must be a certain relation between 
cut-offs of the successive cylinders, which relation can only be determines 
after all conditions are known, but then can be definitely computed and 
plotted for reference in operation. 

So far, in discussing the steam engine, cycles only have been treated 
These cycles are of such a nature that they can be only approached ho 
practice, but since all conclusions have been arrived at through reasoning 
based on assumed laws or hypotheses, the term hypothetical may be appliec 
to all these cycles. It is desirable to compare the actual pressure-volume 
diagram, taken from the indicator card of a steam engine, and the hypothetica 
diagram most nearly corresponding with the conditions. 

In Fig. 106 is shown in full lines a pressure-volume diagram which ha 



WORK OF PISTON ENGINES 



367 



jeen produced from an actual indicator card taken from a simple non-con- 
lensing, four-valve engine having 5 per cent clearance. 

Finding the highest pressure on the admission line A'B' and the lowest 
pressure on the exhaust line D'E', these pressures are regarded as (initial 
pressure) and (back pressure) and a hypothetical diagram constructed cor- 
responding to Cycle III, with cut-off and compression at the same fraction of 
stroke as in the actual engine. 

The first difference between the hypothetical and actual PV diagrams 
is that the point of release C' is not at the end of stroke, as was assumed for the 
hypothetical release, C, a difference which is intentional, since it requires time 
jfor pressure to fall after release to the exhaust pressure. This same fact may 
icause the corner of the diagram to be rounded instead of sharp as at D. 
[Similarly, the point of admission F r is before the end of the return stroke has 



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FIG. 106. Diagram to Illustrate Diagram Factors. 

)een reached, and for a similar reason the corner A' may be rounded, though 
\ release and admission are made sufficiently early the corners D f and A' will be 
harp, approaching the hypothetical, H and A. 

i These differences, however, have little effect upon the area of the actual 
tiagram, which is seen to be much smaller than the hypothetical. This 
jeficiency of area is the net result of a large number of influences, only a few 
I which can be fully explained in connection with the pressure volume 
fccussion. 

[ Beginning with the point of admission, F', the line F'A'B' represents 
|he period of admission. The rounding at A' has been explained; the indina- 
Jon of the line from A' toward B' is due in part to the frictional loss of 
sure as the fluid passes at high velocity through ports and passages from 
am chest to cylinder. As the stroke progresses, the linear velocity of the 
n increases toward mid-stroke, requiring higher velocities in steam 
ages. The greater consequent friction causes pressure to fall in the 
inder. The resistance of pipes and valves leading to the engine have 






368 ENGINEERING THERMODYNAMICS 



an effect on the slope of this line. As cut-off is approached, this pressure 
fall becomes more rapid, due to the partial closure of the admission valve. 
From B', the point of cut-off, to C", the point of release, is the period 
of expansion, during which the pressures are much lower than during the 
hypothetical expansion line BC, due principally to the lower pressure at the 
point of cut-off B' than at B. Hence, the frictional fall in pressure during 
admission has a marked effect upon the work done during expansion. The 
curve B'C' rarely follows the law PV = const, exactly, though it commonly 
gives approximately the same work area. During the first part of expansion, 
the actual pressure commonly falls below that indicated by this curve, but 
rises to or above it before the expansion is complete. This is largely due 
to condensation of steam on the cylinder walls at high pressures, and its 
reevaporation at lower pressure, to be studied in connection with a thermal 
analysis of the cycle. The curve of expansion may also depart from this 
very considerably, due to leakage, either inwardly, through the admission 
yalve, or by piston from a region of higher pressure, or outwardly, through 
exhaust valve, or by piston into a region of lower pressure, or by drain, 
indicator, or relief valves, or by stuffing-boxes. 

From the opening of the exhaust valve at the point of release, C', till 
its closure at compression E f , is the period of exhaust. Pressures during this 
period, as during admission, are affected by frictional losses in the passages 
for steam, in this case the pressure in the cylinder being greater than that 
in exhaust pipe due to friction, by an increasing amount, as the velocity of; 
the piston increases toward mid-stroke. Thus the line DE' rises above the 
line DE until the partial closure of the exhaust near the point of compres- 
sion causes it to rise more rapidly. r ; 

At the point of compression E' the exhaust valve is completely closec 
and the period of compression continues up to admission at F r . Leakage 
condensation, and reevaporation affect this line in much the same way at 
they do the expansion, and often to a more marked degree, due to the facij 
that the volume in cylinder is smaller during compression than during 
expansion, and a given weight condensed, reevaporated, or added or removec 
by leakage will cause a greater change in pressure in the small weigh, 
present than if the change in weight had occurred to a large body of steam! 

In the compound engine all these effects are present in each cylinder iir 
greater or less degree. In addition, there are losses of pressure or of volum 
in the receivers themselves between cylinders, due to friction or conden 
sation, and where especially provided for, reevaporation by means o 
reheating receivers. The effect of these changes in receivers is to cause 
loss of work between cylinders, and to make the discharge volume of on I 
cylinder greater or less than the supply volume of the next, while the? 
were assumed to be equal in the hypothetical cases. 

The effect of all of these differences between the actual and hypothetic^ 
diagrams is to make the actual indicated work of the cylinder something les 
than that represented by the hypothetical diagram. Since these effects ai 



WORK OF PISTON ENGINES 369 

not subject to numerical calculations from data ordinarily obtainable, they 
are commonly represented by a single coefficient or diagram factor which is 
! a ratio, derived from experiment, between the actual work and that indicated 
by hypothesis. 

It is at once evident that there may be more than one hypothetical 
I diagram to which a certain engine performance may be referred as a standard 
I of comparison. When the heat analysis of the steam engine is taken up, 
a standard for comparison will be found there which is of great use. For 
determination of probable mean effective pressure, however, no method of 
calculation has been devised which gives better results than the computa- 
tion of the hypothetical mean effective pressure from one of the standard 
hypothetical diagrams, and multiplying this by a diagram factor obtained 
by experiment from a similar engine, under as nearly the same conditions 
as can be obtained. 

Such diagram factors are frequently tabulated in reference books on the 
steam engine, giving values for the factor for various types and sizes, under 
various conditions of running. Unfortunately, however, the exact standard 
to which these are referred is not stated. In this text it will be assumed, 
unless otherwise stated, that the diagram factor for an actual engine is the 
ratio of the mean effective pressure of the actual engine to that computed 
for Cycle I, without clearance or compression, logarithmic law, with cut-off 
at the same fraction of stroke as usual, initial pressure equal to maximum 
during admission in actual, and back pressure equal to minimum during exhaust 
of the actual engine. 

This is selected as the most convenient standard of comparison for mean 
effective pressures, as it is frequently impossible to ascertain the clearance 
in cases where data are supplied. When it is possible to do so, however, closer 
approximation may be made to the probable performance by comparing 
Ihe actual with that hypothetical diagram most nearly approaching the cycle, 
Using same clearance, cut-off, and compression as are found in the actual. 

Commercial cut-off is a term frequently used to refer to the ratio of the 
volume AK to the displacement, Fig. 106, in which the point K is found on 
E initial pressure line AB, by extending upward from the true point of cut-off 
B' a curve PV = const. 

While the diagram factor represents the ratio of indicated horse-power 
|to hypothetical, the output of power at the shaft or pulley of engine is less 
Irian that indicated in the cylinders, by that amount necessary to overcome 
Mechanical friction among engine parts. If this power output at shaft or 
pulley of engine is termed brake horse-power (B.H.P) then the ratio of this to 
indicated horse-power is called the mechanical efficiency, E m , of the engine 



(583) 



The difference between indicated and shaft horse-power is the power 
imed by friction (F.H.P.). Friction under running conditions consists 



370 



ENGINEERING THERMODYNAMICS 



of two parts, one proportional to load, and the other constant and inde- 
pendent of load, or 

(F.H.P.) =N[(const.) x(m.e.p.) + (const.)2], 
where N is speed, revolutions per minute. But NX (const.) (m.e.p.) = (I.H.P.)^i 



and 



(F.H.P.) = (I.H.P.)^i+A r (const.) 2 , (584) 



where KI and (const.)2 are constants to be determined for the engine, whose 
values will change as the conditions of the engine bearing-surfaces or lubri- 
cation alters. This value for (F.H.P.) may be used to evaluate E m , 



_(I.H.P.)- (F.H.P.) _ _ JV(const.) 2 

~ I.H.P. Al I.H.P. ' ' ' * 



(585) 



but indicated horse-power divided by speed is proportional to mean effective 
pressure, so that 

K 2 



~7~~ \ 

(m.e.p.) 



(586) 



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.80 

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02 1 6 8 10 12 U 16 18 20 22 24 26 28 30 32 31 36 38 10 

Mean Effective Pressure 



FIG. 107. Diagram to Show Relation of Mechanical Efficiency and Mean Effective Pressure. 

From this expression, speed has been eliminated, which agrees with general 
observation, that mechanical efficiency does not vary materially with speed. 
Values of the constants K\ and K% may be ascertained if (m.e.p.) and E m 
are known for two reliable tests covering a sufficient range, by inserting their 
values forming two simultaneous equations. 

The numerical values of KI found in common practice are between .02 
and .05, and for Kz between 1 .3 and 2, in some cases passing out of this range 
In Fig. 107 is shown the form of mechanical efficiency curve when plotted 
on (m.e.p.) as abscissas, using Ki = M, X 2 = 1.6. It may be noted that a1 



WORK OF PISTON ENGINES 371 

ter (m.e.p.) the curve does not approach unity, but the value (1 K\) 
a limit. The mechanical efficiency becomes zero for this case, at a mean 
-ective pressure of about 1.67 Ibs. per square inch, which is that just 
j sufficient to keep the engine running under no load. For a given speed and size 
|)f cylinders, the abscissas may be converted into a scale of indicated 
jiorse-power. 

Prob. 1. Assuming a back pressure of 10 Ibs. per square inch absolute, a clearance 
; jf 8 per cent, a cut-off of 40 per cent, and compression of 20 per cent, show how 
'm.e.p.) varies with initial pressure over a range of 200 Ibs., starting at 25 Ibs. 

Prob. 2. For an initial pressure of 150 Ibs. per square inch absolute, show how 
m.e.p.) varies with back pressure over a range of 30 Ibs., starting at Ib. per square 
toch absolute, keeping other quantities as in Prob. 1. 

Prob. 3. For values of initial pressure, back pressure, etc., as given in Probs. 1 
ind 2, show how (m.e.p.) varies with clearance from 1 per cent to 15 per cent. 

Prob. 4. For values of initial pressure, etc., as given in Probs. 1 arid 2, show how 
pi.e.p.) will vary with cut-off from to 1. 

Prob. 5. For values of initial pressure, etc., as given in Probs. 1 and 2, show how 
'm.e.p.) will vary with compression for values from to 50 per cent. 

Prob. 6. A certain engine developing 675 I.H.P. at a speed of 151 R.P.M., 
[delivered at the shaft 606 H.P. measured by an absorption dynamometer. A second 
Wtet'at 100 R.P.M. gave 150 I.H.P., and 114 shaft H.P. If this engine is to deliver 
[500 H.P. at the shaft at a speed of 150 R.P.M., what will be the I. H.P. and the mechan- 
jical efficiency? 

Prob. 7. A compound Corliss engine, 25 and 52 ins. diameters, 60 ins. stroke, 
double-acting, was designed for 650 I.H P. at 63 R.P.M. It was found that at this 
Aeed and I.H.P. the mechanical efficiency was 91 per cent. When running with no 
load, the cylinders indicated 38.1 I.H.P. at 65 R.P.M. Find the probable mechanical 
ffrciency when developing 300 I.H.P. at a speed of 64 R.P.M. 

24. Consumption of the Steam Engine and its Variation with Valve Move- 
ment and Initial Pressure. Best Cut-off as Affected by Condensation and 
leakage. The weight of steam used by a steam engine per hour divided by 
Ihe indicated horse-power is said to be the water rate or steam consumption 
of that engine. It is almost needless to say that this is not a constant for 
a given engine, since it will change with any change of initial pressure, back 
ftessure, or valve setting, leakage, or temperature conditions. Since there 
ye at least two other uses of terms water rate or consumption, this may be 
ftrmed the actual water rate, or actual consumption, the latter being a more 
general term which may refer to the weight of fluid used per hour per 

licated horse-power, whatever the fluid may be, steam, air, carbon dioxide, 

any other expansive fluid. The present discussion has special reference 

steam. 

From the hypothetical diagram, by computations such as are described 
br the various foregoing cycles, may be obtained a quantity representing 
pe weight of fluid required to develop one horse-power for one hour, by the 



372 ENGINEERING THERMODYNAMICS 

performance of the hypothetical cycle. This may be termed the hypothetico 
consumption or for steam cycles the hypothetical water rate. 

By the use of the actual indicator card, may be obtained, by methods sti] 
to be described, the weight of fluid accounted for by volumes and pressure 
known to exist in the cylinder, this being called the indicated consumption o 
the engine or indicated water rate if the fluid be steam. 

The heat analysis of the steam-engine cycle will lead to another standan 
of comparison which is of the greatest importance as a basis of determining 
how nearly the actual performance approaches the best that could b< 
obtained if the engine were to use all available energy possessed by the steam 
At present the object is to compare the actual and indicated performance wit! 
that hypothetically possible with cylinders of the known size. Accordingly 
attention will be confined first to hypothetical consumption, and the quan title; 
upon which it is dependent. 

For Cycle III, which is the most general for the single-expansion engine 
logarithmic law, the expression for consumption in pounds fluid per hour pei 
indicated horse-power, found in Section 5, Eq. (267), is as follows: 

Hypothetical consumption, Ibs. per hr. per I . H.P. 



in which the value of mean effective pressure itself depends upon (in.pr.) 
(bk.pr.), c, Z, and X. The density of the fluid at initial pressure, d lf is to b 
ascertained from tables of the properties of steam or of whatever fluid is used. 

In Fig. 108 are the results of computations on the hypothetical steam con 
sumption, using mean effective pressures as plotted in Fig. 105. For each curve 
conditions are assumed to be as stated on the face of the diagram, varying onl 
one of the factors at a time. 

Other conditions remaining unchanged, it may be noted that consumptio 
decreases for an increase of initial pressure, though not rapidly in the highe 
pressure range. 

Cut-off has a marked effect upon consumption, the minimum occurring 
when cut-off is such as to give complete expansion. This occurs when 

1+c ___ (in.pr.) 
Z'+c~ (bk.pr.)' 
or 



(in.pr.) 



which may be termed hypothetically best cut-off. In the case assumed in til 
diagram, 



WORK OF PISTON ENGINES 



373 



If clearance be varied, maintaining constant compression and cut-off, large 
trance will give high consumption due to an excessive quantity of fluid 
equired to fill the clearance space. Extremely small clearance leads to a high 
pressure at the end of compression, causing a loss of mean-effective pressure, 
ijid consequent high consumption. Between, the consumption has a minimum 
>oint, which is dependent for its location on both cut-off and compression. 

Decreasing back pressure has a beneficial effect upon mean effective pressure 
,md consumption. This would be still more marked in the figure if a case had 
>een selected with a very short cut-off. 

Compression, throughout the ordinary range of practice, has but slight effect 
ipon consumption, indicated by the flat middle portion of the curve in Fig. 




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40 


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80 


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120 


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INITIAL PRESSURE 100 LBS.ABS 
BACK PRESSURE 15 
CLEARANCE -10 
CUT OFF .25 
COMPRESSION .35 
































































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pack.Pre 


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11 pressi 


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5 10 15 20 25 30 .1 2 .3 .4 .5 .6 .7 .8 .9 1.0 



ho. 108. Curves to Show the Variation of Hypothetical Steam Consumption of Simple 
Engines, Logarithmic Expansion and Compression. 

08. Very small or zero compression permits too much high-pressure steam to 
K admitted to the clearance space without doing work, and excessively large 
learance causes pressures during compression to rise very high, thereby de- 
Teasing mean effective pressure; hence this curve of consumption rises at 
poth ends. 

Hypothetically, the best attainable consumption for given initial and back 
jessures is obtained when both expansion and compression are complete. 
I The indicated consumption, or, as it is frequently called for the steam engine, 
fsteam accounted for by the indicator card" " or " indicated water rate," is 
itermined from the indicator card as follows. Let Fig. 109 represent an indicator 
flagram. The points of cut-of #nd compression are located from the form of 



374 



ENGINEERING THERMODYNAMICS 



the line, at the highest point on the expansion line and the lowest point on th' 
compression line respectively. The fraction of the card lengths completed a 
cut-off, 



AD> 



and the fraction of card length from point of compression to end of stroke, 



are determined, the pressure at cut-off and compression measured by th 
proper vertical scale, and the corresponding densities, BI, and 2 respectively 
are ascertained from steam tables for dry saturated steam. Clearance, Cl, i 
known or ascertained by the form of the compression curve (Chap. I, Section 12 



o 



CutOff 




lAtm. 



B C 



D V 



FIG. 109. Diagram to Illustrate Method of Determining Indicated Water Rate of Sten 

Engine. 
> 

At the point of cut-off, the weight of dry saturated steam present in t 
cylinder is D(Z+c)Bi, and at compression the weight present is D(X+c) 
on the assumption that the steam in the cylinder is of density 81 and 82 
these two instants. Accepting this assumption, the weight of steam used j: 
cycle is 



Wt. steam per cycle 
The work per cycle 



[(Z+c)$i- (X+c)Z 2 ]D^ 



TF=144D(m.e.p.), 
and for n cycles per minute the indicated horse-power is 

T IT P _144n/Xm.e.p.) 
i.n.r --- --.>., ()()n 



375 



_wm_ = 60X33,OOOXD[(Z+c)Bi-(X+c)o 2 ]n 
I. H. P. 144wD(m.e.p.) 



or, 

Ind. consumption, Ibs. per hr. per I.H.P. 



(589 > 



which is the expression used to find indicated consumption for either simple- 
or multiple-expansion engines. In applying this to the multiple-expansion 
engine the terms Z, X and c are found for any one cylinder, and the mean effective 
pressure is referred to that cylinder. There may be, therefore, as many computa- 
tions as there are expansion stages. For a compound engine, for instance, 
indicated consumption according to high-pressure card is found by inserting 
in formula Z, X and c for the high-pressure card, 81 and 82 for corresponding 
pressures, and for (m.e.p.) use 

(m.e.p. ref. to H.P.) = (m.e.p.)^+(m.e.p.) L -^. . . . (590) 

If the computation is done by means of events on the low-pressure card, 
the (m.e.p.) must be referred to the low. 

(m.e.p. ref. to L.P.) = (m.e.p.)/r^+(m.e.p.)z, (591) 

In general for a multiple-expansion engine 

(m.e.p. ref. to cyl. A) = 2(m.e.p.)yc- (592) 

jt// 






It is often difficult and sometimes impossible to determine the point of 

i cut-off and of compression on the indicator card. The expansion and compres- 

sion lines are of very nearly hyperbolic in form and are usually recognizable. 

The highest point on the hyperbolic portion of the expansion line is regarded as 

cut-off, and the lowest point on the hyperbolic portion of the compression line, 

as the point of compression. It must be understood that by reason of the 

[condensation and re-evaporation of steam in cylinders the weight of steam 

>roper is not constant throughout the stroke, so that calculations like the 

ibove will give different values for every different pair of points chosen. The 

lost correct results are obtained when steam is just dry and these points are 

it release and compression most nearly. 



376 ENGINEERING THERMODYNAMICS 

When under test of actual engines the steam used is condensed and weighed 
and the indicated horse-power determined, then the actual steam consumption 
or water rate can be found by dividing the weight of water used per hour in 
the form of steam by the indicated horse-power. This actual water rate is 
always greater than the water rate computed from the equation for indicated 
consumption. The reasons for the difference have been traced to (a) leakage 
in the engine, whereby steam weighed has not performed its share of work, to 
(b) initial condensation, whereby steam supplied became water before it could 
do any work, (c) variations in the water content of the steam by evaporation 
or condensation during the cycle, whereby the expansion and compression laws 
vary in unpredictable ways, affecting the work. 

Estimation of probable water rate or steam consumption of engines cannot, 
therefore, be made with precision except for engines similar to those which have 
been tested, in all the essential factors, including, of course, their condition, 
and for which the deficiencies between actual and indicated consumptions 
have been determined. This difference is termed the missing water, and end- 
less values for it have been found by experiments, but no value is of any use 
except when it is found as a function of the essential variable conditions that 
cause it. No one has as yet found these variables which fix the form for an 
empiric formula for missing water nor the constants which would make such a 
formula useful, though some earnest attempts have been made. This is nc 
criticism of the students of the problem, but proof of its elusive nature, and the 
reason is probably to be found in the utter impossibility of expressing by 
formula the leakage of an engine in unknown condition, or the effect of its 
condition and local situation on involuntary steam condensation and evapora-i 
tion. It is well, however, to review some of these attempts to evaluate J 
missing water so that steam consumption of engines may be estimated 
After studying the many tests, especially those of Willans, Perry announcer 
the following for non-condensing engines, in which the expansion is but littlt 



Missing water Z 
Indicated steam J - / ~^ T) ' 



where d is the diameter of the cylinder in inches and TV the number of revolution 
per minute. This indicates that the missing steam or missing water has bee 
found to increase with the amount of expansion and decrease with diameter c 
cylinder and the speed. Thermal and leakage conditions are met by the us 
of difference values of m, for there are given 

m = 5 for well-jacketed, well-drained cylinders of good construction wit 
four poppet valves, that is, with minimum leakage and condensatior 

w = 30 or more for badly drained un jacketed engines with slid, valves, 
is, with high leakage and condensation possibilities. 

m = 15 in average cases. 



WORK OF PISTON ENGINES 377 

For condensing engines Perry introduces another variable the initial pres- 
ij sure pounds per square inch absolute, p giving 



Missing water = __ _ _ 
Indicated steam 



It might seem as if such rules as these were useless, but they are not, especially 

when a given engine or line of engines is being studied or two different engines 

I compared; in such cases actual conditions are being analyzed rather than 

j predictions made, and the analysis will always permit later prediction of con- 

siderable exactness, if the constants are fixed in a formula of the right empiric 

form. Similar study by Heck has resulted in a different formula involving 

different variables and constants, but all on the assumption that the dis- 

crepancies are due to initial condensation. He proposes an expression equi- 

valent to 



Missing steam _ .27 IS(x2xi) 
Indicated steam ~~ ^/]y\ piZ ' 



in which ]V = R.P.M. of the engine; 
d = diameter in inches; 
L = stroke in feet; 

S = the ratio of cylinder displacement surface in square feet to dis- 
placement in cubic feet. 



L d ' 



The term (x 2 -Xi) is a constant supposed to take into account the amount 
Srf initial condensation dependable on the difference between cylinder wall 
knd live-steam temperature and is to be taken from a table found by trial as 
Ihe difference between the x for the high pressure and x for the low pressure, 
both absolute, see Table XIV at the end of the Chapter. 

In discussing the hypothetical diagrams, it was found that best economy was 
[.obtained with a cut-off which gives complete expansion. For other than 
hypothetical diagrams this is not true, which may be explained most easily by 
inference to the curves of indicated, and actual consumption, and missing 
(jBteam, Fig. 110. 

The curve ABC is the hypothetical consumption or water rate for a certain 
steam engine. Its point of best economy occurs at such a cut-off, B, that expan- 



378 



ENGINEERING THERMODYNAMICS 



sion is complete. The curve GUI is computed by Heck's formula for missing 
water. The curve falls off for greater cut-offs. Adding ordinates of these tw< 
curves, the curve DEF for probable consumption is found. The minimun 
point in this curve, E, corresponds to a longer cut-off than that of ABC 
Since cut-off B gave complete expansion, cut-off E must give incomplete expan 
sion. In other words, due to missing steam, the condition which really give 
least steam consumption per hour per indicated horse-power corresponds to j 
release pressure, which is greater than the back pressure. 

It should be noted that the minimum point mentioned above will not b< 
best cut-off, for the output of the engine is not indicated, but brake horse-power 



30 

1 

S80 

1 

10 




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N 


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C 


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) .1 .2 .a 

Per Cent Cut off 






FIG. 110. Diagram to Show Displacement of Best Cut-off Due to Effect of Missing Wate 
from Point B for the Hypothetical Cycle to Some Greater Value E. 

In Fig. Ill on cut-off as abscissa are plotted (EFG) consumption pound 
per hour per I.H.P., and for the case assumed, (OD) the curve of mechanic; 
efficiency, based on cut-off. 

(Ibs. steam per hr.) . B.H.P. _ (Ibs. steam per hr.) 
I.H.P. *I.H.P.~ B.H.R ' 

or, in other words, 

Consumption, Ibs. per hr. per I.H.P. 

-- - = Consumption, Ibs. per hr. per B.H.P. (59('i 



Due to the increasing value of E m for greater cut-offs, the minimum point 
corresponds to a cut-off still longer than for the curve EFG, which itself W 
found hi Fig. 110 to give a longer cut-off than that of the hypothetical curve 

Hence the best cut-off for economy of steam, where the net power at tl 
shaft is regarded as the output, will be such as to give incomplete expansion 
or a release pressure above back pressure, this effect being caused by bot 
missing steam and by frictional losses. 

Prediction of actual consumption of steam engines as a general propositk: 
is almost hopeless if any degree of accuracy worth while is desired, though tl 



WORK OF PISTON ENGINES 



379 



effect on steam consumption of changing the value of any one variable can be 
pretty well determined by the previous discussion qualitatively, that is, in kind, ' 
though not quantitatively in amount. Probably the best attempt is that 
of Hrabak in German, which takes the form of a large number of tables 
developed from actual tests though not for engines of every class. These tables 
are quite extensive, being in fact published as a separate book and any abstrac- 
tion is of no value. 

There is, however, a sort of case of steam consumption prediction that can be 
carried out with surprising precision and that is for the series of sizes or line 
of engines manufactured by one establishment all of one class, each with 
about the same class of workmanship and degree of fit, and hence having 
leakage and cylinder condensation characteristics that vary consistently through- 
out the whole range. For such as these tables and curves of missing water 




o .1 .2 .3 

Per Cent Cut off 
FIG. 111. Diagram to Show Relation of Steam per Hour per I.H.P. and per B.H.P. to Cut-off. 

can be made up and by the best builders are, for making guarantees of steam 
consumption for any service conditions that their engines are able to meet. 

The practice of one firm making what is probably the best line of stationary 
engines in this country is of sufficient interest to warrant description. The 
primary data are curves of indicated water rate plotted to mean effective 
pressure for clearances of three or four per cent, and that mean effective pressure 
is chosen in any one specific case that will give the horse-power desired at the 
fixed speed for some one set of cylinder sizes avaliable. To this indicated 
water rate a quantity is added constituting the missing water which is made 
up of several parts as follows: The first is an addition representing condensa- 
tion which is plotted in curve form as a function of (a) boiler pressure, (6) 
superheat in the steam, (c) piston speed, (d) the class of engine simple, com- 
pound or triple, with jacketed or unjacketed cylinders, and for cylinder ratios 
from 4 to 1, to 6 to 1 in the case of compounds. It is therefore a most complex 
quantity, the nature of the variations in which can only be indicated here. 



380 ENGINEERING THERMODYNAMICS 

For example, increase of piston speed decreases the condensation loss as 
does multiple expansion, and also jacketing, while increase of superheat in the 
steam also decreases it, but superheat has less effect in triple than in com- 
pounds and less in compound than in simple engines. 

The next factor of correction is that covering leakage losses, also additive 
to indicated water rate and which with it and the condensation loss make 
up the probable steam consumption. The leakage decreases regularly with 
increase of piston speed, is less for large than for small engines, the change 
being rather fast from 50 to 200 horse-power and much slower later, being 
scarcely anything at all over 2000 horse-power. 

Example 1. What cut-off will give the lowest indicated water rate for a 9x12- 
in. engine, with 5 per cent clearance and no compression when running non-condensing 
on an initial pressure of 100 Ibs. per square inch gage, and what will be the value 
of this water rate? What steam will be used per hour per brake horse-power 
hypothetically? From Eq. (587) 



= (1+.05)- -.05=8.7 per cent, 
115 

and 

(m.e.p.) =115|.087X(.087+.05) log^fl -15=27.2 Ibs. sq.in. 
L .lo7J 

Hence 

Steam per hour per I.H.P. = ~- - .137-. 05 X^ X. 262 = 17.2 Ibs. 

2il ,2i [ lloj 

From the curve of Fig. 107, assuming it to apply to the engine, for this value of 
(m.e.p.) mechanical efficiency is 90 per cent, hence from Eq. (596) the weight of 
steam per shaft horse-power per hour will be 19.1 pounds. 

Prob. 1. Draw diagram similar to Fig. 108 for following case: 

Initial pressure, 135 Ibs. per square inch gage, back pressure 10 Ibs. per square 
inch absolute, clearance 5 per cent, cut-off 30 per cent, compression 25 per cent. 

Prob. 2. From indicator diagram shown in Fig. 106 find the indicated water rate 
of the engine from which it was taken. 

Prob. 3. The indicated water rate of a 9xl2-in. jacketed engine when running 
non-condensing at a speed of 250 R.P.M. with an initial pressure of 100 Ibs. per 
square inch gage and \ cut-off is 50 Ibs. Using Perry's formula what will be the 
probable actual steam used by engine per horse-power hour " 

Prob. 4. A 24 x48-in. engine in good condition is found to have an indicated water 
rate of 25 Ibs. when cut-off is i, initial pressure 100 Ibs. per square inch gage, back 
pressure 10 Ibs. per square inch absolute, and speed of 125 R.P.M. What will be the 
missing water, and the rate as found by Perry's formula and by Heck's? 



WORK OF PISTON ENGINES 381 

Prob. 5. What will be the probable amount of steam used per hour by a 36x48- 
in. engine with 5 per cent clearance running at 100 R.P.M. on an initial pressure of 
150 Ibs. per square inch gage a back pressure of 5 Ibs. per square inch absolute, \ cut- 
off and 10 per cent compression? 

Prob. 6. How will the amount of steam of Prob. 5 compare with that used by 
a 15X22x36-in engine with 5 per cent clearance in each cylinder, running at 100 
R.P.M. on same pressure range with J cut-off in high-pressure cylinder, \ cut-off in 
low, and 10 per cent compression in each cylinder? 



25. Variation of Steam Consumption with Engine Load. The Willans 
Line. Most Economical Load for More than One Engine and Best Load 
Division. However valuable it may be to the user of steam engines to have 
an engine that is extremely economical at its best load which, it should be 
noted, may have any relation to its rated, horse-power, it is more important 
usually that the form of the economy load curve should be as flat as possible 
and always is this case when the engine must operate under a wide range of 
load. This being the case it is important to examine the real performance 
curves of some typical engines all of which have certain characteristic 
similarities as well as differences. 

From the discussion of hypothetical and indicated water rates it appears 
that the curve of steam consumption (vertical) to engine load (horizontal) 
is always concave upward and always has a minimum point, not at the maxi- 
| mum load. Actual consumption curves are similar in general form, but as 
has been pointed out, the load at which the water rate is least corresponds 
to some greater mean effective pressure than that for the hypothetical, so 
the whole curve is displaced upwards and to the right by reason of cylinder 
condensation and leakage losses. This displacement may be so great as to 
prevent the curve rising again beyond the minimum point, in which case the 
least steam consumption corresponds to the greatest load. Just what form 
(the actual water rate-load curve will take depends largely on the form of valve 
gear and type of governing method in use, by throttling initial pressure with 
a fixed cut-off or, by varying cut-off without changing initial pressure, with or 
without corresponding changes in the other valve periods. 

Whenever the control of power is by throttling of the supply steam the 
curve is found to be almost exactly an hyperbola, so that (water rateXhorse- 
Jpower) plotted to horse-power is a straight line which being characteristic 
| is much used in practical work and is known as the Willans line. All other 
| engines, that is, those that govern on the cut-off, have Willans lines that 
j are nearly straight, such curvature as exists being expressed by a second 
i degree equation instead of one of the first degree. 

Equations for Willans lines can always be found for the working range 
jtof load, that is, from about half to full load, though not for the entire range, 
fccept in unusual cases, and these equations are of very great value in pre- 
dicting the best division of load between units, which is a fundamental step 
[In deciding, how many and what sizes of engine to use in carrying a given 
load in industrial power plants. 



382 



ENGINEERING THERMODYNAMICS 




\\ 



\ 



m 




o o 

5* o5 

'd *H 'I Joel rawo^s jo '{ 



WORK OF PISTON ENGINES 



3S3 



Before taking up the derivation of equations some actual test curves will 
>e examined and a number of these are grouped in Fig. 112 for engines of 
i-arious sizes, simple and compound, up to 10,000 H.P., on which vertical 
distances represent pounds of steam per hour, per I.H.P. and horizontal 
.H.P. To show the essential similarity of the curves for engines of different 
ize more clearly, these are re-plotted in Fig. 113 to a new load scale based 
;>n best load of each, which is taken as unity. This is evidently a function of 
{lean effective pressure, just what sort of function does not matter here. In 




75 loo 125 

Percentage of most economical load 



150 



IG. 113. Typical Water Rate Lead Curves for Steam Engines Plotted to Fractional Loads. 

ery case the Willans line is also plotted in Fig. 112, each line being num- 
3red to correspond to its water rate curve. 

[ As there is a corresponding similarity of form for the water rate and 
fcllans line of steam turbines, though the reasons for it will be developed 
fcr, it must be understood that the mathematical analysis that follows applies 
both turbine and piston steam engines, and finally it makes no difference 
jiat units are used for load, whether I.H.P., or B.H.P. or K.W. of a direct- 
mnectecl electric generator. 

In Fig. 114 is shown the water-rate curve to a K.W. base for the 10,000 K.W. 
Irtis steam turbine at the Chicago Edison, Fiske Street Station for which the 
llowing equation fits exactly: 



384 



ENGINEERING THERMODYNAMICS 



where Y = pounds of steam per hour -r- 1000; 

P = load (in this case in K.W.H1000; 

Y 

p = pounds of steam per K.W. hr. 

A similar equation fits fairly well the curve of Fig. 115, representing tl 
7000 H.P. piston engines of the Interborough Railway, Fifty-ninth Strei 
station, as well as the combined piston engine and low-pressure steam turbii 



400 




4oOO 5000 



7000 9000 11000 
1000 P 5= Kilowatts 



13000 



15000 



FIG. 114. Performance of a 10000-K.W. Steam Turbine. 



taking its exhaust steam, in the same station, but with different numeric 
constants, as below: 

Y 36 7^ 
Piston engine, -p = ^ 



Combined piston engine and turbine, B = ~T>' -- 2.90+.713P. 

A third case of smaller size is shown in Fig. 116, representing the p 
formance of a 1000-K.W. Corliss piston engine driving a generator for wh 
the equation is 



WORK OF PISTON ENGINES 

Kilowatts 



385 




7000 8000 9000 10000 11000 12000 13000 14000 15000 16000 
Kilowatts =1000 P 

FIG. 115. Performance of a 7000-H.P. Piston Engine alone and with a Low-pressure 

Steam Turbine. 































































































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25 24 




































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-S 15 ^ 18 

g 


















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K) 








a 


K) 








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Kilowatts = 1000 P 
FIG. 116. Performance of a 1000-K.W. Steam Turbine. 



386 ENGINEERING THERMODYNAMICS 

These illustrations could be multiplied indefinitely, but those given will 
suffice to establish the fact that the two following equations are fundamental 
over the working range of any steam engine of whatever type: 

Water rate line, j=~+B+CP . ........ (597) 

Water per hour, Willans line, Y = A+BP+CP 2 , . . (598 

in which F is the weight of steam per hour and P the engine load whether 
expressed in indicated or brake horse-power, or in kilowatts. 

At the most economical load the water rate is a minimum, so that 

A 
dP 

whence the most economical load is 

U 

(599) 

Where the Willans line is straight, C = 0, and the most economical load 
is the greatest load. 

Two engines carrying the same load must divide it and some one pro- 
portion may be best. To find out, consider first any number of similar engines, 
that is, engines that have the same constants A, B, and C, denoting each case 
by subscripts. Then 

Let P = total load; 
" PI, P 2 , P 3 , etc. = individual engine loads; 

F = total water per hour; 
" YI, F 2 , F 3 = water per hour for each engine. 

Then 

.+ F n 

. Pn) + C(F 
-P 3 2 + . . .+P n 2 ). 

Only the last term is variable and this is a minimum when 



Therefore for similar engines, the best division of load is an- equal division. 



WORK OF PISTON ENGINES 387 

When the engines are dissimilar it is convenient to first consider the case 
straight Willans lines for which C = 0. Then for two such engines 

Y = A 1 +A 2 +B 1 P 1 +B 2 P 2 

= (A l +A 2 )+B l (P-P 2 )+B 2 P 2 
= (A l +A 2 )+B 1 P+(B 2 -B 1 )P 2 . 

t any given load P the first two terms together will be constant, and the 
r per hour will be least when the last term is least. As neither factor 
i be zero, this will occur when P 2 is least. 

Therefore for two dissimilar engines the best division of load is that which 
ts the greatest possible share on the one with the smaller value of B, in its equation, 
yvided each has a straight Willans line. 
Two dissimilar engines of whatever characteristics yield the equation, 

Y = A 1 +A 2 +B l P l +B 2 P 2 +CiP i 2 +C 2 P 2 2 
= (Ai+A 2 +B ] P+C 1 P 2 ) 

+ (B 2 -2PC 1 -B l )P 2 

+(Ci+C 2 )P 2 2 . 

Differentiation with respect to P 2 , and solving for P 2 , the load for the second 
gine that makes the whole steam consumption least, gives, 



(600) 



= constant + constant X P. 



Therefore, the load division must be linear and Eq. (600) gives the numerical 
ue, when any two engines share a given load. 

This sort of analysis can be carried much further by those interested, but 
tee forbids any extension here. It is proper to point out, however, that 
means of it the proper switch-in points for each unit in a large power station 
L be accurately found, to give most economical operation on an increasing 
tion load. 

i 26. Graphical Solution of Problems on Horse-power and Cylinder Sizes . 

je diagram for mean effective pressures in terms of initial and back 
(assure, clearance, compression and cut-off, Fig. 117, facilitates the solution 
1 3q. (262) in Section 5. The mean effective pressure is the difference 
\reen mean forward and mean back pressure. The former is dependent 
i clearance, cut-off and initial pressure. In the example shown on the 
fe by letters and dotted lines, clearance is assumed 5 per cent, shown at 
Project horizontally to the point F, on the contour line for the assumed 
pff, 12 per cent. Project downward to the logarithmic scale for " mean 



388 



ENGINEERING THERMODYNAMICS 




'I 



o 

-> 



s 

i 

I 

Q 
1 

^3 



WORK OF PISTON ENGINES 389 

ard pressure in terms of initial pressure " to the point G. On the scale 
" initial pressure " find the point H, representing the assumed initial pres- 
ure, 115 Ibs. absolute. Through -G and H a straight line is passed to the point 
C on the scale for " mean forward pressure/' where the value is read, 
i.f.p. = 49.5 Ibs. absolute. 

Mean back pressure is similarly dependent upon clearance, compression 
iid back pressure, and the same process is followed out by the points A, B, 
1, D and E, reading the mean back pressure, 3.2 Ibs. absolute at the point E. 
?hen by subtraction, 

(m.e.p.) = (m.f. p.) -(m.b.p.) =49.5-3.2 = 46.3 Ibs. 

Fig. 118 is arranged to show what conditions must be fulfilled in order to 
f<btain equal work with complete expansion in both cylinders in a compound 
ngine, finite receiver, logarithmic law, no clearance, Cycle VII, when low- 
ressure admission and high-pressure exhaust are not simultaneous. This is 
ecussed in Section 11, and the diagram represents graphically the conditions 
xpressed in Eqs. (376), (377), (378), (379). 

To illustrate its use assume that in an engine operating on such a cycle, 
le volume of receiver is 1.5 times the high-pressure displacement, 1.5 = 2/, then 

=.667. Locate the point A on the scale at bottom of Fig. 118, corresponding 

o this value. Project upward to the curve marked "ratio of cut-offs " and at 
le side, C, read ratio of cut-offs 

= .572. 



text extending the line AB to its intersection D, with the curve GH, the point 
is found. From D project horizontally to the contour line representing the 
iven ratio of initial to back pressure. In this case, initial pressure is assumed 
in times back pressure. Thus the point E is located. Directly above E at 
|ie top of the sheet is read the cylinder ratio, at F, 



If cylinder ratio and initial and final pressures are the fundamental data of 
problem, the ratio of cut-offs and ratio of high-pressure displacements 
receiver volume may be found by reversing the order. 



390 



ENGINEERING THERMODYNAMICS 




WORK OF PISTON ENGINES 391 



GENERAL PROBLEMS ON CHAPTER III. 

Prob. 1. How much steam will be required to run a 14xl8-in. double-acting 
engine with no clearance at a speed of 200 R.P.M. when initial pressure is 100 Ibs. 
per square in. gage, back pressure 28 ins. of mercury (barometer reading 30 ins ), and 
cut-off is ? What will be horse-power under these conditions? 

NOTE: 5 for 100 Ibs. =.26, for 28 ins. =.0029. 

Prob. 2. Draw the indicator cards and combined diagram for a compound steam 
engine without receiver, and with 3 per cent clearance in low pressure and 5 per cent in 
high, when initial pressure is 100 Ibs. per square inch absolute, back pressure 10 Ibs. 
per square inch absolute, high-pressure cut-off , high-pressure compression yV, and 
low pressure compression -5. 

Prob. 3. A simple double-acting engine, 18x24 ins., is running at 100 R.P.M. 
on compressed air, the gage pressure of which is 80 Ibs. The exhaust is to atmosphere. 
If the clearance is 6 per cent and cut-off f , and compression 10 per cent, what horse- 
power is being developed, the expansion being adiabatic, and how long can engine be 
run at rated load on 1000 cu.ft. of the compressed air? 

Prob. 4. Will the work be equally distributed in a 12xl8x24-in. engine with 
infinite receiver and no clearance when cut-off is | in high pressure cylinder, and f in 
low, expansion being logarithmic, initial pressure 150 Ibs. per square inch absolute 
and back pressure atmosphere? What will be work in each cylinder? 

Prob. 5. The receiver of a 15X20x22 in. engine is 4 times as large as high- 
pressure cylinder. What will be the horse-power, steam used per hour, and variation 
in receiver pressure for this engine, if clearance be considered, zero and initial pressure 
is 125 jbs. per square inch gage, back pressure 5 Ibs. per square inch absolute, cut-offs 
and f in high- and low-pressure cylinders respectively, and piston speed is 550 ft. per 
minute? 

NOTE: 8 for 125 Ibs. =.315, for 5 Ibs. =.014. 

Prob. 6. At no load an engine having 7 per cent clearance, cuts off at 4 per cent 
of its stroke, while at full load it cuts off at 65 per cent of its stroke. At no load, 
compression is 40 per cent and at full load 5 per cent. What percentage of full-load 
horse-power is required to overcome friction, and what percentage of steam used at 
full load, is used on friction load, if initial pressure is constant at 100 Ibs. per square 
inch gage, back pressure constant at 5 Ibs. per square inch absolute, and expansion 
is logarithmic? 

NOTE: 8 for 100 Ibs. =.262, for 5 Ibs. =.014. 

Prob. 7. The initial pressure on which engine is to run is 115 Ibs. per square inch 
gage, and steam is superheated and known to give a value of s = 1.3. For an engine 
in which clearance may be neglected, work is to be equal, and expansion complete 
in both cylinders, when back pressure is 10 Ibs. per square inch absolute. What must 
be the cut-offs and cylinder ratio to accomplish this when receiver is 3 times high 
pressure cylinder volume? 

Prob. 8. A 12-in. and 18x24 ins. double-acting engine with zero clearance and 
infinite receiver operates on an initial pressure of 150 Ibs. per square inch gage, and 



392 ENGINEEEING THERMODYNAMICS 

a back pressure of 5 Ibs. per square inch absolute. What will be the release and receiver 
pressures, horse-power, and steam consumption when speed is 150 R.P M. ; expansion 
logarithmic, and cut-off | in each cylinder? 

NOTE: 8 for 150 Ibs. = .367, for 5 Ibs. = .014. 

Prob. 9. If a third cylinder 24 ins. in diameter were added to engine of Prob. 8 
and cut-off in this made , how would horse-power, steam consumption, receiver and 
release pressures change? 

Prob. 10. What would have to be size of a single cylinder to give same horse-power 
at same revolutions and piston speed as that of engine of Prob. 8 under same conditions 
of pressure and cut-off? 

Prob. 11. With the higii-pressure cut-off at \, and low and intermediate cut-offs 
at fV, what will be the horse-power, water rate and receiver pressures of a 30 X 48 X 77 X72- 
in. engine running at 102 R.P.M. on an initial pressure of 175 Ibs. per square inch gage 
and a back pressure of 26 *ins. of mercury (barometer reading 30 ins.), if the 
receiver be considered infinite and expansion logarithmic, clearance zero? What change 
in intermediate and low-pressure cut-offs would be required to give equal work distribu- 
tion? 

NOTE: 8 for 175 Ibs. =.419, for 26 ins. =.0058, 

Prob. 12. If it had been intended to have all the cut-offs of the engine of Prob. 
11, equal to ^, what should have been the size of the intermediate and low-pressure 
cylinders to give equal work for same pressure range and same high-pressure cylinder? 

Prob. 13. To attain complete expansion in all cylinders of the engine of Prob. 11, 
with the initial and back pressures as there given, what cut-offs would be required 
and what receiver pressures would result? 

Prob. 14. A compound locomotive has no receiver, the high-pressure clearance 
is 8 per cent, and low-pressure clearance 5 per cent. The cylinders are 22 and 
33 X48 ins., high-pressure cut-off f , high- and low-pressure compression each 10 per 
cent, initial pressure 175 Ibs. per square inch gage, back pressure one atmosphere, and 
expansion and compression logarithmic. What will be the horse-power at a speed 
of 40 miles per hour, the engine having 7-ft. driving wheels? At this speed, how 
long will a tank capacity of 45,000 gallons last? 

NOTE: 5 for 175 Ibs. =.419, for 15 Ibs. =.038. 

Prob. 15. A superheater has been installed on engine of Prob. 14 and expansion 
and compression, now follow the law PV S =c, when s = 1.2. What effect will this have 
on the horse-power and steam consumption? 

Prob. 16. What will be the maximum receiver pressure work done in each cylinder 
and total work for a cross-compound engine 36 and 66 X48 ins., running at 100 R.P.M. 
on compressed air of 100 Ibs. per square inch gage pressure, exhausting to atmosphere 
if the high pressure cut-off is , clearance 6 per cent, compression 20 per" cent, low- 
pressure cut-off is f, clearance 4 per cent, compression 15 per cent, and receiver volume 
is 105 cu.ft.? 

Prob. 17. A manufacturer gives the horse-power of a 42x64x60-in. engine as 
2020, when run at 70 R.P.M. on an initial pressure of 110 Ibs. per square inch gage, 
atmospheric back pressure, and .4 cut-off in high-pressure cylinder. How does this 
value compare with that found on assumption of 5 per cent clearance in high, 4 per 
cent in low, and complete expansion and compression is each cylinder? 

Prob. 18. A mine hoisting engine is operated on compressed air with a pressure of 
150 Ibs. per square inch absolute and exhausts to atmosphere. The cylinder sizes are 
26X48X36 ins., and clearance is 5 per cent in each. At the start the high-pressure 



WORK OF PISTON ENGINES 393 

t-off is 1 and low pressure $, while normally both cut-offs are . The exhaust 
from high-pressure cylinder is into a large receiver which may be considered infinite. 
] The compression is zero at all times. Considering the exponent of expansion to be 
i 1.4, what will be the horse-power under the two conditions of cut-off given, for a speed 
of 100 R.P.M.? 

Prob. 19. What must be ratio of cylinders in the case of a compound engine with 
: infinite receiver, to give equal work distribution complete expansion and com- 
j pression if the least clearance which may be attained is 5 per cent in the high- 
pressure cylinder, and 3 per cent in the low-pressure. The engine is to run non- 
; condensing on an initial pressure of 125 Ibs. per square inch gage, with expansion 
exponent equal to 1.3? What must be the cut-offs and compressions to satisfy these 
j conditions? 

Prob. 20. Assuming 7 per cent clearance in high-pressure cylinder and 5 per cent 
in low, infinite receiver, and no compression, how will the manufacturer's rating of 
2100 H.P. check, for a 36X6x48-in, engine running at 85 R.P.M. on an initial 
pressure of 110 Ibs. per square inch gage, and a back pressure of 26 ins. vacuum, with 
.3 cut-off in high pressure cylinder? 

Prob. 21. For a 25X40x36-in. engine, with 5 per cent clearance, I cut-off 
and 20 per cent compression in each cylinder, what will be horse-power for an initial 
pressure of 100 Ibs. per square inch gage, and a back pressure of 17.5 Ibs. per square inch 
absolute, with logarithmic expansion and compression? 

Prob. 22. What must be the cylinder ratio and cut-off to give complete expansion 
ki a no-clearance, 14 and 22 x24-in. engine with no receiver and logarithmic expansion, 
when initial pressure is 100 Ibs. per square inch gage, and back pressure 10 Ibs. per 
square inch absolute? What will be the horse-power and steam used for these conditions 
at a speed of 150 R.P.M.? 

NOTE: 5 for 100 Ibs. =.262, for 10 Ibs. =.026. 

Prob. 23. A 24X20x24 in. engine with no receiver or clearance, runs on com- 
pressed air of 120 Ibs. per square inch gage pressure, and exhausts to atmosphere. 
When running at a speed of 125 R.P.M. , with high-pressure cut-off \, what horse- 
power will be developed and how many cubic feet of compressed air per minute will 
be required to run the engine, the expansion being adiabatic? Will the work be equally 
divided between the two cylinders? 

Prob. 24. It is desired to run the above engine as economically as possible. What 
change in' cut-off will be required, and will this cause a decrease or increase in horse- 
power and how much? How will the quantity of air needed be affected? 

Prob. 25. A mill operates a cross-compound engine with a receiver 3 times as largo 
Iks high- pressure cylinder, on an initial pressure of 125 Ibs. per square inch gage, and a 
lack pressure of 10 Ibs. per square inch absolute. The engine may be considered as 
Beithout clearance, and the expansion as logarithmic. As normally run the cut-off in 
[thigh-pressure cylinder is f and in low, J. It has been found that steam is worth 25 
I cents a thousand pounds. What must be charged per horse-power day (10 hours) 
||to pay for steam if the missing water follows Heck's formula? 
NOTE 8 for 125 =.315, for 10 = .026. 

Prob. 26. By installing a superheater the value of s in Prob. 25 could be changed 
1.3. The cost of st am would then be 30 cents a thousand pounds. From the effect 
value of s alone would the installation of the superheater pay? 

Prob. 27. When a 26X48x36-in. cross-compound engine with a receiver volume 
35 cu.ft. and zero clearance, is being operated on steam of 125 Ibs. per square inch 



394 ENGINEERING THEEMODYNAMICS 

gage initial pressure, and atmospheric exhaust, is the work distribution equal, wht 
high-pressure cut-off is f and low-pressure cut-off |? For these cut-offs what 
fluctuation in receiver pressure and what steam will be used per horse-power hour? 

NOTE: 8 for 125 = .315, for 15 lbs.=.038. 

Prob. 28. To operate engine of Prob. 27 under most economical conditions, wh; 
values must be given to the cut-offs, and what values will result for receiver pressure 
horse-power, and .steam used per hour? 

Prob. 29. What will be the horse-power and steam used by a 20x30x364 
engine with infinite receiver and no clearance, if expansion be such, that s = 1.2 
high-pressure cut-off f ,. low-pressure cut-off |, initial pressure 100 Ibs. per square inc 
gage, back pressure 3 Ibs. per square inch absolute, and speed 100 R.P.M. 

NOTE: S for 100 Ibs. =.262, for 3 Ibs. = .0085. 

Prob. 30. The following engine with infinite receiver and no-clearance, runs c 
steam which expands according to the logarithmic law. Cylinders 9, and 13 x: 
ins., initial pressure 125 Ibs. per square inch gage, back pressure 5 Ibs. per square in< 
absolute, high-pressure cut-off , low-pressure }, speed 150 R.P.M. What will t 
horse-power and steam consumption hypothetical and probable? 

NOTE: S for 125 Ibs. =.315, for 5 Ibs. =.014. 

Prob. 31. By graphical means find the (m.e.p.), of a 15X22x30-in. crof: 
compound engine, with 5 per cent clearance in each cylinder, if the receiver volume 
8 cu.ft., initial pressure 125 Ibs. per square inch absolute, back pressure 10 Ibs. p 
square inch absolute, high-pressure cut-off f , low-pressure &, high-pressure comprt 
sion 40 per cent, low-pressure 20 per cent, high-pressure crank following 90, logarithm 
expansion. 

Prob. 32. Show by a series of curves, assuming necessary data, the effect < 
(m.e.p.) of cut-off, back pressure, clearance, and compression. 

Prob. 33. Show by curves, how the indicated, and actual water rate, of an 18 X2: 
in. engine with 5 per cent clearance, and running at 125 R.P.M. on an initial presst 
of 125 Ibs. per square inch gage, and a back pressure of 10 Ibs. per square inch absolu 
may be expected to vary with cut-off from ^o to f . 



TABLES 



395 



TABLE XIII 



PISTON POSITIONS FOR ANY CRANK ANGLE 

FROM BEGINNING OF STROKE AWAY FROM CRANK SHAFT TO FIND PISTON POSITION FROM 
DEAD-CENTER MULTIPLY STROKE BY TABULAR QUANTITY 



Crank 
Angle. 


-U 

r 


-U 5 

r 


5 

r 


f 5 .5 


f 6 


i* 

r 


fs 


-'=9 

T 


5 


.0014 


.0015 


.0015 


.0016 


.0016 


.0016 


.0017 


.0019 


10 


.0057 


.0059 


.0061 


.0062 


.0063 


.0065 


.0067 


.0076 


15 


.0128 


.0133 


.0137 


.0140 


.0142 


.0146 


.0149 


.0170 


20 ! .0228 


.0237 


.0243 


.0248 


.0253 


.0260 


.0265 


.0302 


25 


.0357 


.0368 


.0379 


.0388 


.0394 


.0405 


.0413 


.0468 


30 


.0513 


.0531 


.0545 


.0556 


.0565 


.0581 


.0592 


.0670 


35 


.0698 


.0721 


.0740 


.0754 


.0767 


.0787 


.0801 


.0904 


40 


.0910 


.0939 


.0962 


.0981 


.0997 


.1022 


.1041 


.1170 


45 


.1152 


.1187 


.1215 


.1237 


.1256 


.1286 


.1308 


.1468 


50 


.1416 


.1458 


.1491 


.1518 


.1541 


.1576 


.1607 


.1786 


55 


.1713 


.1759 


. 1828" 


.1827 


.1853 


.1892 


.1922 


.2132 


60 


.2026 


.2079 


.2122 


.2157 


.2186 


.2231 


.2295 


.2500 


65 


.2374 


.2431 


.2477 


.2514 


.2545 


.2594 


.2630 


.2886 


70 


.2730 


.2794 


.2844 


.2885 


.2929 


.2973 


.3013 


.3290 


75 


.3123 


.3187 


.3239 


.3282 


.3317 


.3372 


.3414 


.3705 


80 


.3516 


.3586 


.3642 


.3687 


.3725 


.3784 


.3828 


.4132 


85 


.3944 


.4013 


.4068 


.4113 


.4151 


.4210 


.4254 


.4564 


90 


.4365 


.4437 


.4495 


.4547 


.4580 


:4641 


.4686 


.5000 


95 


.4816 


.4885 


.4940 


, .4985 


.5022 


.5081 


.5126 


.5436 


100 


.5253 


.5323 


.5378 


.5424 


.5461 


.5520 


.5564 


.5868 


105 


.5711 


.5775 


.5828 


.5870 


.5905 


.5961 


.6002 


.6294 


110 


.6150 


.6214 


.6265 


.6306 


.6340 


.6393 


.6530 


.6710 


115 


.6600 


.6657 


.6703 


.6740 


.6771 


.6820 


.6856 


.7113 


120 


.7026 


.7080 


.7122 


.7157 


.7186 


.7231 


.7265 


.7500 


125 


.7449 


.7495 


.7533 


.7563 


.7588 


.7628 


.7658 


.7868 


130 


.7844 


.7885 


.7920 


.7947 


.7969 


.8004 


.8030 


.8214 


135 


.8223 


.8258 


.8286 


.8308 


.8327 


.8357 


.8379 


.8535 


140 


.8570 


.8600 


.8623 


.8642 


.8658 


.8682 


.8703 


.8830 


145 


.8889 


.8913 


.8931 


.8946 


.8958 


.8978 


.8993 


.9096 


150 


.9173 


.9191 


.9204 


.9216 


.9226 


.9241 


.9252 


.9330 


155 


.9420 


.9432 


.9452 


.9451 


.9457 


.9468 


.9476 


.9531 


160 


.9625 


.9633 


.9640 


.9645 


.9650 


.9656 


.9661 


.9698 


165 


.9787 


.9792 


.9796 


.9799 


.9802 


.9805 


.9809 


.9829 


170 


.9905 


.9908 


.9909 


.9911 


.9912 


.9913 


.9915 


.9924 


175 


.9976 


.9977 


.9977 


.9977 


.9978 


.9978 


.9979 


.9981 


180 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 


1.0000 



396 



ENGINEERING THERMODYNAMICS 



TABLE XIV 
VALUES OF x FOR USE IN HECK'S FORMULA FOR MISSING WATER 



Absolute 
Steam Pressure. 


X 


Absolute 
Steam Pressure. 


X 


Absolute 
Steam Pressure. 


X 





170 


70 


297.5 


165 


393 


1 


175 


75 


304 


170 


397 


2 


179 


80 


310 


180 


405 


3 


183 


85 


316 


185 


409 


4 


186 


90 


321.5 


190 


413 


6 


191 


95 


327 


K5 


416.5 


8 


196 


100 


332.5 


200 


420 


10 


200 


105 


338 


210 


427 


15 


210 


110 


343 


220 


431 


20 


220 


115 


348 


230 


441 


25 


229 


120 


353 


240 


447.5 


30 


238 


125 


358 


250 


454 


35 


246 


130 


362.5 


260 


460.5 


40 


254 


135 


367 


270 


467 


45 


262 


140 


371.5 


280 


473 


50 


269.5 


145 


376 


290 


479 


55 


277 


150 


380.5 


300 


485 


60 


284 


155 


385 






65 


291 


160 


389 







TABLE XV 
SOME ACTUAL ENGINE DIMENSIONS 

SIMPLE 



7X9 


7|X15 


16 X18 


15|X24 


24 X3G 


8X9 


8|X15 


16|X18 


16 X24 


26 X36 


9X9 


12 X15 


17 X18 


18 X24 


26|X38 


5|X10 


13 X15 


17|X18 


20 X24 


28 X36 


6^X10 


14 X15 


18 X18 


22 X24 


14 X42 


8 X10 14|X15 


19 X18 


24 X24 


15 X42 


9 X10 


15 X15 


20 X18 


16|X27 


16 X42 


10 X10 


16 X15 


29 X19 


17^X27 


18 X42 


11 X10 


17^X15 


12 X20 


10 X30 


20 X42 


9|X10| 


11 X16 


14 X20 


12 X30 


22 X->2 


10X10 


12 X16 


18 X20 


16 X30 


24 X42 


7JX12 


13 X16 


19 X20 


18 X30 


26 X42 


8 X12 


14iXl6 


28 X20 


18fX30 


28 X^ 


8fX12 


15 X16 


21 X20 


20 X30 . . 


18 X48 


9 X12 


15iX16 


22 X20 


24 X30 


20 X48 


10 X12 


16 X16 


12 X21 


22 X33 


22 X48 


11 X12 


17 X16 


13 X21 


24 X33 


24 X48 


1UX12 


18 X16 


18|X21 


10 X36. 


26 X48 


12 X12 


18JX17 


20 X21 


12 X36 


28 X48 


12|X12 


23 X17 


20 X22 


14 X36 


24 X54 


13 X12 


26 X17 


18 X24 


16 X36 


26 X54 


14 X12 


10 X18 


10 X24 


18 X36 


28 X54 


10 X14 


11 X18 


12 X24 


20 X36 


28 X60 


11 X14 


15 X18 


14|X24 


22 X36 





TABLES 



397 



TABLE XV. Continued 



COMPOUND 

NOTES: 1 to run condensing or non-condensing on initial pressure of 100-150. 

2 to run condensing or non-condensing on initial pressure of 100. 

3 to run condensing or non-condensing on initial pressure of 125. 

4 to run condensing or non-condensing on initial pressure of 90-100. 

5 to run condensing or non-condensing on initial pressure of 110-130. 

6 to run condensing or non-condensing on initial pressure of 140-160. 

7 to run condensing or non-condensing on initial pressure of 125 



4J- 8 X 6 




13 -18 X14 


1 


11 -19 X18 


3 


14i-26 X22 


4 


22 -38 X33 


3 


6 -10 X 6 




13 -20 X14 


1,2 


12 -21 X18 


3 


18 -32 X22 


5 


24 -42 X33 


3 


7 -13 X 8 




7f-13| X 15 


3 


13 -22^X18 


3 


10 -17^X24 


3 


181-32^X36 


3 


6 -12 X10 


1 


9 -151X15 


3 


14 -24 X18 


1 


11 -19 X24 


3 


20 -36 X36 


3 


7 -12 X10 


1 


11 -19 X15 


4 


151-26^X18 


3 


12 -18 X24 


7 


26^-46 X36 


3 


8 -12 X10 


1 


13 -19 X15 


5 


16 -24 X18 


1 


13 -20 X24 


7 


28^-50 X36 


3 


8f-15|X10 




7|-13-|X16 


3 


16 -26 X18 




14 -22 X24 


7 


14-25 X42 


3 


7 -14 X10 


1 


9 -15|X16 


3 


16^-28^X18 


3 


16^-28|X24 


3 


151-26^X42 


3 


8 -14 X10 


1 


10 -17I-X16 


3 


8 -12 X20 


7 


17|-30|X24 


3 


18^-32^X42 


3 


9|-15 Xll 


2 


11 -19 X16 


3 


9 -14 X20 


7 


22 -38 X24 


3 


20 -36 X42 


3 


7|-13|X12 


3 


11 -22 X16 


1 


16 -28 X20 


1 


24 -42 X24 


3 


16|-28|X48 


3 


9 -151X12 


3 


12 -21 X16 


3 


17 -30 X20 


1 


12 -21 X27 


3 


17^-30^X48 


3 


19 -14 X12 


1 


13 -22 X16 


1 


18 -28 X20 


1 


13 -22|X27 


3 


22 -38 X48 


3 


10 -16 X12 


1 


13 -221X16 


3 


19 -30 X20 


1 


16^-28^X27 


3 


; 24 -42 X48 


3 


10 -18 X12 


1 


14 -22 X16 


2 


19 -30 X22 


1 


171-30^X27 


3 


! 181-32^X54 


3 


11 -16 X12 


1 


14J-25 X16 


3 


9 -15|X21 


3 


141-25 X30 


3 


20 -36 X54 


3 


9 -18 X14 


1 


15 -22 X16 


- 


12 -21 X21 


3 


151-26|X30 


3 


26 -46 X54 


3 


10 -18 X14 


1 


15|-26|X16 


3 


13 -22|X21 


3 


18 1 -32|X30 


3 


28^-50 X54 


3 


10 -17^X14 


3 


16 -25 X16 


2 


14i-26|X21 


3 


20 -36 X30 


3 


22 -38 X60 


3 


11 -19 X14 


3 


13 -23 X17 


4,6 


151-28^X21 


3 


28^-50 X30 


3 


24 -42 X60 


3 


11 -18 X14 


1 


15 -26 X17 


4,6 


18f-32! X 21 


3 


30 -54 X30 


3 


30 -54 X60 


3 


2 -18 X14 


1 


16|-29 X17 


4 


20 -36 X21 


3 


161-28^X33 


3 


32|-57 X60 


3 


12 -20 X14 


1 


9 -15|X18 


3 


13 -23 X22 


5 


17^X30^X33 


3 


34 -60 X60 


3 



TRIPLE 

NOTE: All condensing and to run of initial pressure as given. 



SIZE. 


P 


SIZE. 


P 


SIZE. 


P 


10 -151-26X15 
11-18 -30X20 


200 
250 


27 -43 { SJX39 
\ oi 


180 


30-50-82X48 
25-41-68X48 


180 
190 


12 -20 -34X24 


180 


18 -28M8X40 


180 


27-45-75X54 


190 


12 -19 -32X24 


190 


22 -37 -63X42 


180 


28-45-72X54 


185 




175 


22 -38 -64X42 


185 


28-46-75X54 


180 


12-2-22 -36X24 
14 -23 -28X26 


180 
190 


32|-53 -{^X48 


265 


29-47-83X54 
32-52-92X50 


160 
200 


; 18 -29 -47X30 
16|-24 -41X30 


200 
180 


35-57-1^X48 


265 


34-56-100X60 
35-58 (~oX 60 


200 
190 


18 -30 -50X30 
16 -25-1-43X30 


200 
190 


36 -57 - {^X48 


295 


\ 69 
34-57-104X63 


200 


16|-24 -41X30 


180 


28 -45 -72X48 


180 







CHAPTER IV 

HEAT AND MATTER. QUALITATIVE AND QUANTITATIVE RELATIONS 
BETWEEN THE HEAT CONTENT OF SUBSTANCES AND THEIR PHYSICAL- 
CHEMICAL STATE, 

1. Substances and Heat Effects Important in Engineering. It has been 
shown in preceding chapters concerned with work in general and with the deter- 
mination of quantity of work that may be done in the cylinders by, or on 
expansive fluids that 

(a) Fluids originally at low may be put in a high-pressure condition by 
the expenditure of work; 

(b) Fluids under high pressure may do work in losing that pressure. 

That work may be done, fluids under pressure are necessary and that the 
greatest amount of work may be done per unit of fluid the fluid itself must be 
expansive, that is, it must be a gas or a vapor. Gases or vapors under pressure 
are, therefore, prerequisites to the economical use of fluids for the doing of work, 
and that this work may be done at the expense of heat or derived from heat, it 
is only necessary that the heat be used to create the necessary primary con- 
dition of high pressure in vapors and gases. There are two general ways of 
accomplishing this purpose first, to apply the heat to a boiler supplied with 
liquid and discharging its vapor at any pressure as high as desired or as high 
as may be convenient to manage; second, to apply the heat to a gas confined 
in a chamber, raising its pressure if the chamber be kept at a fixed volume, 
which is an intermittent process, or increasing the fluid volume if the size of the 
chamber be allowed to increase, the fluid pressure being kept constant or not, 
and this latter process may be intermittently or continuously carried out. 

These two processes are fundamental to the steam and gas engines that are 
the characteristic prime movers or power generators of engineering practice, 
utilizing heat energy, and with the exception of water-wheels the sole commer- 
cially useful sources of power of the industrial world. Thus, the heating of 
gases and the evaporation of liquids are two most important thermal processes 
to be examined together with their inverse, cooling and condensation, and 
necessarily associated in practical apparatus with the heating and cooling of 
solid containers or associated liquids. From the power standpoint, the effects 
of heat on solids, liquids, gases and vapors, both without change of state and with 
change of state are fundamental, and the substances to be studied as heat carriers 
do not include the whole known chemical world, but only those that are cheap 
enough to be used in engineering practice or otherwise essential thereto. These 
substances of supreme importance are, of course, air and water, with all their 

398 



HEAT AND MATTER 399 

jiysical and chemical variations, next the fuels, coal, wood, oil, alcohol and 
jmbustible gases, together with the chemical elements entering into them 
| id the chemical compounds which mixed together may constitute them. 
] Probably next in importance from the standpoint of engineering practice 
je the substances and thermal processes entering into mechanical refriger- 
jion and ice making. There are but three substances of commercial importance 
jsre ammonia, pure and in dilute aqueous solution, carbonic acid and air. 
iae process of heating or cooling solids, liquids, gases and vapors, together 
Hth solidification of water into ice, evaporation and condensation, fundamental 
f power problems, are also of equal importance here, but there is added an 
ilditional process of absorption of ammonia vapor in water and its discharge 
om the aqueous solution. 

Many are the practical applications of heat transfer or transmission, some 
which call into play other "substances than those named. In the heating 
| buildings there is first combustion with transfer of heat to wa,ter in boilers, 
i)W of the hot water or steam produced to radiators and then a transfer of heat 
^ the air of the room; in feed-water heaters, heat of exhaust steam warms 
ater on its way to. the boilers; in economizers, heat of hot flue gases is trans- 
rred to boiler feed water; in steam superheaters, heat of hot flue gases is trans- 
rred to steam previously made, to raise its temperature, steam pipes, boiler 
lirfaces and engine cylinders transfer heat of steam to the air which is opposed 
|r covering and lagging, in steam engine condensers heat of exhaust steam is 
;ansf erred to circulation water; in cooling cold storage rooms and making ice, 
i solution of calcium or sodium chloride in water is circulated through pipes 
jid tanks and is itself kept cool in brine coolers in which the brine transfers 
he heat absorbed in the rooms and tanks, to the primary substance ammonia 
F carbonic acid and evaporates it. 

While evaporation and condensation as processes are fundamental to the 
tachinery and apparatus of both power and refrigeration, they also are of 
aportance in certain other industrial fields. In the concentration of 
>lutions to promote crystallization such, for example, as sugar, evapora- 
m of the solution and condensation of the distillate are primary processes 
[fclso is the case in making gasolene and kerosene from crude oil, in the making 
ftalcohol from a mash, and many other cases found principally in chemical 
jianufacture. These are examples of evaporation and condensation in which 
jttle or no gases are present with the vapor but there are other cases in which 
! gas is present in large proportion, the thermal characteristics of which are 
lifferent as will be seen later. Among these processes are: the humidification 
; t moistening of air with water in houses and factories to prevent excessive 
lin evaporation of persons breathing the air, excessive shrinkage of wood-work 
to facilitate the manufacturing processes like tobacco working and thread 
finning. Conversely, air may be too moist for the purpose, in which case it 
fcried by cooling it and precipitating it$ moisture as rain or freezing it out as 
m, and this is practiced in the Gayley process of operating blast furnaces, where 
Mbess of moisture will on dissociating absorb heat of coke combustion and reduce 



400 ENGINEERING THERMODYNAMICS 

the iron output per ton of coke, and in the factories where, for example, collodior 
is worked, as in tb~ manufacture of photographic films, with which moisture 
seriously interferes. Of course, humidification of air by water is accomplishec 
only by evaporation of water, and evaporation of water is only to be accomplishec 
, by the absorption of heat, so that humidification of air by blowing it over watei 
or spraying water into it must of necessity cool the water, and this is the prin 
ciple of the cooling tower or cooling pond for keeping down the temperature 
of condenser circulating water, and likewise the principle of the evaporative 
condenser, in which water cooler and steam condenser are combined in one 
The same process then, may serve to cool water if that is what is wanted, or tt 
moisten air, when dry air is harmful, and may also serve to remove moistun 
from solids like sand, crystals, fabrics, vegetable or animal matter to be reducee 
to a dryer or a pulverized state. 

There are some important examples of humidification in which the substance; 
are not air and water, and one of these is the humidification of air by gasolene 
or alcohol vapor to secure explosive mixtures for operating gas engines. Her< 
the air vaporizes enough of the fuel, humidifying or carburetting itself to serv< 
the purpose, sometimes without heat being specifically added and sometime, 
with assistance from the hot exhaust. A somewhat similar action takes plac< 
in the manufacture of carburetted water gas when the water gas having n< 
illuminating value is led to a hot brick checkerwork chamber supplied with i 
hydrocarbon oil, the vapor of which humidifies the gas, the heat of vaporizatioi 
being supplied by the hot walls and regularly renewed as the process is inter 
mittent. Of course, in this case some of the vapors may really decompos 
into fixed gases, a peculiar property of the hydrocarbon fuels, both liquid an* 
gaseous, and frequently leaving residues of tar, or soot, or both. 

Finally, among the important processes there is to be noted that of gasifica 
tion of solid and liquid fuels in gas producers and vaporizers, a process alsj 
carried on in blast furnaces in which it is only an accidental accompanimen 
and not the primary process. Some of the actions taking place in gas producei 
are also common to the manufacture of coal gas, and coke, in retorts, beehiv 
and by-product ovens. 

From what has been said it should be apparent that engineers are conceme 
not with any speculations concerning the nature of heat but only with the kin 
and quantity of effect that heat addition to, or abstraction from, substance 
may be able to produce and not for all substances either. While this interee 
is more or less closely related to philosophic inquiry, having for its object tb 
development of all embracing generalizations or laws of nature, and to tl 
relation of heat to the chemical and physical constitution of matter, subje< 
matter of physical chemistry, the differences are marked, and a clearly define 
field of application of laws to the solution of numerical problems dealing wit 
identical processes constitutes the field of engineering thermodynamics. 

It is not possible or desirable to take up and separately treat every sing 
engineering problem that may rise, but on the contrary to employ the scientil 
methods of grouping thermal processes or substance effects into types. 



HEAT AND MATTER 401 

Prob. 1. Water is forced by a pump through a feed-water heater and economizer 

> a boiler where it is changed to steam, which in turn passes through a superheater 

) a cylinder from which it is exhausted to a condenser. Which pieces of apparatus 

lave to do with heat effects and which with work? Point out similarities and 

jifferences of process. 

Prob. 2. Air is passed over gasolene in a carburetter; the mixture is compressed, 
jurned and allowed to expand in a gas engine cylinder. Which of the above steps have 
h do with heat effects and which with work effects? 

Prob. 3. In certain types of ice machines liquid ammonia is allowed to evaporate, 
ic vapor which is formed being compressed and condensed again to liquid. Which of 
if|ese steps is a work phase and which a heat phase? Compare with Problem 1. 
|> Prob. 4. When a gun is fired what is the heat phase and what is the work 
hase? Are they separate or coincident? 

Prob. 6. Air is compressed in one cylinder, then it is cooled and compressed to 
igher pressure and forced into a tank. The air in the tank cools down by giving 
p heat to the atmosphere. From the tank it passes through a pipe line to a heater 
ad then to an engine from which it is exhausted to the atmosphere. Which steps 
:| the cycle may be regarded as heat and which as work phases? Compare with 
'roblem 2. 

2. Classification of Heating Processes. Heat Addition and Abstraction 
rith, or without Temperature Change. Qualitative Relations. That heat 

pass from a hot to a less hot body if it gets a chance is axiomatic, so that a 
>ody acquiring heat may be within range of a hotter one, the connection between 
inem being, either immediate, that is they touch each other, or another body 
nay connect them acting as a heat carrier, or they may be remote with no more 
lovable connection than the hypothetic ether as is the case with the sun and 
rth. A body may gain heat in other ways than by transfer from a hotter 
>ody, for example, the passage of electrical current through a conductor will 
eat it, the rubbing of two solids together will heat both or perhaps melt one, 
lie churning of a liquid will heat it, the mixing of water and sulphuric acid will 
reduce a hotter liquid than either of the components before mixture, the absorp- 
ion by water of ammonia gas will heat the liquid. All these and many other 
imilar examples that might be cited have been proved by careful investigation 
>artly experimental, and partly by calculation based on various hypotheses to 
>e examples of transformation of energy, mechanical, electrical, chemical, into 
le heat form. While, therefore, bodies may acquire heat in a great many 
different concrete ways they all fall under two useful divisions: 

(a) By transfer from a hotter body; 

(6) By transformation into heat of some other energy manifestation. 

One body may be said to be hotter than another when it feels so to the 
iense of touch, provided neither is too hot or too cold for injury to the tissues, 
Ir more generally, when by contact one takes heat from the other. Thus. 
Seas of heat can scarcely be divorced from conceptions of temperature and the 
definition of one will involve the other. As a matter of fact temperature as 
indicated by any instrument is merely an arbitrary number located by some- 
lody on a scale, which is attached to a substance on which heat has some visible 



402 ENGINEEEING THEEMODYNAMICS 

effect. Temperature is then a purely arbitrary, though generally acceptec 
number indicating some heat content condition on a scale, two points of whicii 
have been fixed at some other conditions of heat content, and the scale spac 
between, divided as convenient. Examination of heat effects qualitative! 
will show how thermometers might be made or heat measured in terms of an 
handy effect, and will also indicate what is likely to happen to any substanc 
when it receives or loses heat. Some of the more common heat effects of variou 
degrees of importance in engineering work are given below: 

Expansion of Free Solids. Addition of heat to free solids will cause them t l 
expand, increasing lengths and volumes. Railroad rails and bridges are longe 
in summer than winter and the sunny side of a building becomes a little highe 
than the shady side. Steam pipes are longer and boilers bigger hot, than cole 
and the inner shell of brick chimneys must be free from the outer to permi 
it to grow when hot without cracking the outer or main supporting stack bodj 
Shafts running hot through lack of lubrication or overloading in comparative! 
cool bearing boxes may be gripped tight enough to twist off the shaft or merel; 
score the bearing. 

Stressing of Restrained Solids. A solid being heated may be restraine 
in its tendency to expand, in which case there will be set up stresses in the mate 
rial which may cause rupture. Just as with mechanically applied loads, bodie 
deform in proportion to stress up to elastic limit, as stated by Hooke's law, s 
if when being heated the tendency to expand be restrained the amount .c 
deformation that has been prevented determines the stress. A steam pip 
rigidly fixed at two points when cold will act as a long column in compressio 
and buckle when hot, the buckling probably causing a leak or rupture. If fixe 
hot, it will tend to shorten on cooling and being restrained will break something 
Cylinders of gas engines and air compressors are generally jacketed with wate 
and becoming hot inside, remaining cold outside, the inner skin of the mete 
tends to expand while the outer skin does not. One part is, therefore, in tensio 
and the other in compression, often causing cracks when care in designing i 
not taken and sometimes in spite of care in large gas engines. 

Expansion of Free Liquids. Heating of liquids will cause them to expan 
just as do solids, increasing their volume. Thus, alcohol or mercury in gla^ 
tubes will expand and as these liquids expand more than the glass, a tube whic 
was originally full will overflow when hot, or a tube of very small bore attache 
to a bulb of cold liquid will on heating receive some liquid; the movement c 
liquid in the tube if proportional to the heat received will serve as a thermometei 
If the solid containing the liquid, expanded to the same degree as the liqui 
there would be no movement. Two parts of the same liquid mass may b 
unequally hot and the hotter having expanded will weigh less per cubic foo 
that is, be of less density. Because of freedom of movement in liquids the lightc 
hot parts will rise and the cooler heavy parts fall, thus setting up a circulatior 
the principle of which is used in hot water heating systems, the hot water froi 
the furnace rising to the top of the house through one pipe and cooling on 11 
downward path through radiators and return pipe. In general then, liquie 



HEAT AND MATTER 403 

decrease in density on heating and increase in density on cooling, but a most 
important exception is water, which has a point of maximum density just 
above the freezing-point, and if cooled below this becomes not heavier but lighter. 
Consequently, water to be cooled most rapidly should be cooled at first from 
the top and after reaching this point of maximum density, from the bottom, if it 
is to be frozen. 

Rise of Pressure in Confined Liquids. When liquids are restrained from 
expanding under heating they suffer a rise of pressure which may burst the 
containing vessel. For this reason, hot water heating systems have at their 
highest point, open tanks, called expansion tanks, which contain more water 
when the system is hot than when cold, all pipes, radiators and furnaces being 
constantly full of water. Should this tank be shut off when the water is cold 
something would burst, or joints leak, before it became very hot. 

Expansion of Free Gases. Just as solids and liquids when free expand under 
heating, so also do gases and on this principle chimneys and house ventilation 
systems are designed. The hot gases in a chimney weigh less per cubic foot 
than cooler atmospheric air; they, therefore, float as does a ship on water, 
the superior density of the water or cold gas causing it to flow under and 
lift the ship or hot gas, respectively. Similarly, hot-air house furnaces and 
ventilating systems having vertical flues supplied with hot air can send it upward 
by simply allowing cold air to flow in below and in turn being heated flow up 
and be replaced. 

Rise of Pressure in Confined Gases. Gases when restrained from expanding 
under heat reception will increase in pressure just as do liquids, only over greater 
ranges, and as does the internal stress increase in solids when heated under 
restraint. It is just this principle which lies at the root of the operation of 
guns and gas engines. Confined gases are rapidly heated by explosive combus- 
tion and the pressure is thus raised sufficiently to drive projectiles or pistons 
in their cylinders. 

Melting of Solids. It has been stated that solids on being heated expand 
but it should be noted that this action cannot proceed indefinitely. Continued 
heating at proper temperatures will cause any solid to melt or fuse, and the pre- 
viously rising temperature will become constant during this change of state. 
IThus, melting or fusion is a process involving a change of state from solid 
to liquid and takes place at constant temperature. The tanks or cans of ice- 
making plants containing ice and water in all proportions retain the same 
temperature until all the water becomes ice, provided there is a stirring or cir- 
[ culation so that one part communicates freely with the rest and provided also 
the water is pure and contains no salt in solution. Impure substances, such as 
I liquid solutions, may suffer a change of temperature at fusion or solidification. 
; For pure substances, melting and freezing, or fusion and solidification, are 
| constant temperature heat effects, involving changes from solid to liquid, or 
liquid to solid states. 

Boiling of Liquids. Ebullition. Continued heating of solids causes fusion, 
and similarly continued heating of liquids causes boiling, or change of state from 



404 ENGINEERING THERMODYNAMICS 

liquid to vapor, another constant temperature process just what temperature, 
will depend on the pressure at the time. So constant and convenient is this 
temperature pressure relation, that the altitude of high mountains can be found 
from the temperature at which water boils. The abstraction of heat from a 
vapor will not cool it, but on the contrary cause condensation. Steam boilers 
and ammonia refrigerating coils and coolers are examples of evaporating appara- 
tus, and house heating radiators and steam and ammonia condensers of con- 
densing apparatus. 

Evaporation of Liquids; Humidification of Gases. When dry winds blow 
over water they take up moisture in the vapor form by evaporation at any 
temperature. This sort of evaporation then must be distinguished from ebul- 
lition and is really a heat effect, for without heat being added, liquid cannot 
change into vapor; some of the necessary heat may be supplied by the water 
and some by the air. This process is general between gases and liquids and is 
the active principle of cooling towers, carburetters, driers of solids like wood 
kilns. The chilling of gases that carry vapors causes these to condense in part. 
As a matter of fact it is not necessary for a gas to come into contact to produce 
this sort of evaporation from a liquid, for if the liquid be placed in a vacuum 
some will evaporate, and the pressure finally attained which depends on the tem- 
perature, is the vapor pressure or vapor tension of the substance, and the amount 
that will so evaporate is measured by this pressure and by the rate of removal 
of that which formed previously. 

Evaporation of Solids. Sublimation. Evaporation, it has been shown, may 
take place from a liquid at any temperature, but it may also take place directly 
from the solid, as ice will evaporate directly to vapor either in the presence 
of a gas or alone. Ice placed in a vacuum will evaporate until the vapor tension 
is reached, and it is interesting to note that the pressure of vapors above theirij 
solids is not necessarily the same as above their liquids at the same temperature, 
though they merge at the freezing-point. This is the case with ice-water- 
water vapor. 

Change of Viscosity. Heating of liquids may have another effect measured 
by their tendency to flow, or their viscosity. Thus, a thick oil will flow easier 
when heated, and so also will any liquid. If, therefore, the time for a giver | 
quantity to flow through a standard orifice under a given head or pressure bt| 
measured, this time, which is the measure of viscosity, will be less for any liquk 
hot, than cold, for the same liquid. Viscosity then decreases with heat additior 
and temperature rise. 

Dissociation of Gases. When gases not simple are heated and the heatinj 
continued to very high temperatures, they will split up into their elements o 
perhaps into other compound gases. This may be called decomposition or, better 
dissociation, and is another heat effect. Thus, the hydrocarbon C2H 4 wil 
split up with solid carbon soot C and the other hydrocarbon CH4 and steal) 
H2O into hydrogen and oxygen. This is not a constant temperature process 
but the per cent dissociated increases as the temperature rises. 

Dissociation of Liquids. Similar to the dissociation of gases receiving heat a 



HEAT AND MATTER 405 

high temperature is the decomposition of some liquids in the liquid state, notably 
the fuel and lubricating oils, or hydrocarbons which are compounds of H and C 
in various proportions, each having different properties. Sometimes these 
changes of H and C groupings from the old to the new compounds under the 
influence of heating will be at constant and at other times at varying tempera- 
tures; sometimes the resulting substances remain liquid and sometimes soot 
or C separates out, and this is one of the causes for the dark color of some 
cylinder oils. 

Absorption of Gases in Liquids. Liquids will absorb some gases quite freely; 
thus, water will absorb very large quantities of ammonia, forming aqua ammonia. 
Addition of heat will drive off this gas so that another heat effect is the expul- 
sion of gases in solution. Use is made of this industrially in the absorption 
system of ammonia refrigeration. 

Solubility of Solids in Liquids. The heating of liquids will also affect their 
solubility for solid salts; thus, a saturated solution of brine will deposit crystals 
on heat abstraction and take them back into solution on heat addition. 
Certain scale-forming compounds are thrown down on heating the water in- 
tended for boilers, a fact that is made use of in feed-water heating purifiers; 
for these salts increase of temperature reduces solubility. In general then heat 
addition affects the solubility of liquids for solid salts. 

Chemical Reaction. Combustion. If oxygen and hydrogen, or oxygen and 
carbon, be heated in contact, they will in time attain an ignition temperature at 
which a chemical reaction will take place with heat liberation called combus- 
tion, and which is an exothermic or heat-freeing reaction. Another and 
different sort of reaction will take place if C02 and carbon be heated together, 
for these will together form a combustible gas, CO, under a continuation of heat 
reception. This is an endothermic or heat-absorbing reaction. Neither of 
these will take place until by heat addition the reaction temperature, called 
ignition temperature for combustion, has been reached. 

Electrical and Magnetic Effects. Two metals joined together at two separate 
points, one of which is kept cool and the other heated, will be found to carry 
an electric current or constitute a thermo-electric couple. Any conductor 
carrying an electric current will on changing temperature suffer a change of 
resistance so that with constant voltage more or less current will flow; this is 
a second electrical heat effect and like the former is useful only in instru- 
ments indicating temperature condition. A fixed magnet will suffer a change 
of magnetism on heating so that heat may cause magnetic as well as electric 
effects. 

These heat effects on substances as well as some others of not so great engi- 
neering importance may be classified or grouped for further study in a variety 
of ways, each serving some more or less useful purpose. 

Reversible and Non-reversible Processes. There may be reversible and 
non-reversible thermal processes, when the process may or may not be con- 
sidered constantly in a state of equilibrium. For example, as heat is applied 
to boiling water there is a continuous generation of vapor in proportion to the 



406 ENGINEERING THERMODYNAMICS 

heat received; if at any instant the heat application be stopped the evapo- 
ration will cease and if the flow of heat be reversed by abstraction, condensa- 
tion will take place, indicating a state of thermal equilibrium in which the 
effect of the process follows constantly the direction of heat flow and is con- 
stantly proportional to the amount of heat numerically, and in sign, of direction. 
As an example of non-reversible processes none is better than combustion, in 
which the chemical substances receive heat with proportional temperature rise 
until chemical reaction sets in, at which time the reception of heat has no fur- 
ther relation to the temperatures, because of the liberation of heat by com- 
bustion which proceeds of itself and which cannot be reversed by heat 
abstraction. Even though a vigorous heat abstraction at a rate greater than 
it is freed by combustion may stop combustion or put the fire out, no amount 
of heat abstraction or cooling will cause the combined substances to change 
back into the original ones as they existed before combustion. The effect of 
heat in such cases as this is, therefore, non-reversible. 

Constant and Variable Volume or Density. When gases, liquids or solids 
are heated they expand except when prevented forcibly from so doing, and as 
a consequence they suffer a reduction of density with the increase of volume; 
this is, of course, also true of changing liquids to their vapors. It should be 
noted that all such changes of volume against any resistance whatever, occur 
with corresponding performance of some work, so that some thermal processes 
may directly result in the doing of work. Heating accompanied by no volume 
change and during which restraints are applied to keep the volume invariable, 
cannot do any work or suffer any change of density, but always results in change 
of pressure in liquids, gases and vapors and in a corresponding change of internal 
stress in solids. 

Constant and Variable Temperature Processes. Another useful division, and 
that most valuable in the calculation of relations between heat effect and heat 
quantity, recognizes that some of the heating processes and, of course, cooling, 
occur at constant temperature and others with changing temperature. For 
example, the changes of state from liquid to solid, and solid to liquid, or freezing 
and fusion, are constant temperature processes in which, no matter how much 
heat is supplied or abstracted, the temperature of the substance changing state is 
not affected, and the same is true of ebullition and condensation, or the changing 
of state from liquid to vapor, and vapor to liquid. These latter constant- 
temperature processes must not be confused with evaporation, which may 
proceed from either the solid or liquid state at any temperature whether constant 
or not. 

Prob. 1. From the time a fire is lighted under a cold boiler to the time steam 
first comes off, what heat effects take place? 

Prob. 2. What heat effects take place when a piece of ice, the temperature of 
which is 20 F., is thrown onto a piece of red-hot iron? 

Prob. 3. What heat effects must occur before a drop of water may be evaporated 
from the ocean, and fed back into it as snow? 

Prob. 4. What heat changes take place when soot is formed from coal or oil? 



HEAT AND MATTER 407 

Prob. 5. In a gas producer, coal is burned to C0 2 , which is then reduced to CO. 
team is also fed to the producer, and H and O formed from it. Give all the heat 
lects which occur. 

Prob. 6. By means of what heat effects have you measured temperature changes, 
r have known them to be measured? 

Prob. 7. When the temperature changes from 40 F. to 20 F., give a list of all 
eat effects you know that commonly occur for several common substances. Do the 
une for a change in the reverse direction. 

Prob. 8. If a closed cylinder be filled with water it will burst if the temperature 
je lowered or raised sufficiently. What thermal steps occur in each case? 
; Prob. 9. If salt water be lowered sufficiently in temperature, a cake of fresh ice 
jad a rich salt solution will be formed. State the steps or heat effects which occur 
tiring the process. 



'homson " every kind 



408 ENGINEERING THERMODYMAMICS 

of a double metallic bar, often brass and iron, consisting of a piece of e* 
fastened to the other to form a continuous strip. The two metals are expam 
by the temperature different amounts causing the strip to bend under heati 
There are also in use electric forms for all temperatures, and these are 
only reliable ones for high temperatures, both of the couple and resistai 
types except one dependent on the color of a high temperature body, blj 
when cold. That most useful and common class involving the inter 
pendence of pressure and temperature, or volume and temperature, of a fl 
is generally found in the form of a glass bulb or its equivalent, to wh 
is attached a long, narrow glass tube or stem which may be open or closed 
the end; open when the changes of fluid volume at constant pressure are 
be observed and closed when changes of contained fluid pressure at const; 
restrained volume are to be measured as the effect of temperature chan 
For the fluid there is used most commonly a liquid alone such as mercury, 
a gas alone such as air; though a gas may be introduced above mercury i 
there may be used a liquid with its vapor above. When the fluid is a liqi 
such as mercury, in the common thermometer, the stem is closed at the end 
that the mercury is enclosed in a constant-volume container or as nearly so 
the expansion or deformation of the glass will permit, which is not filled w 
mercury, but in which a space in the stem is left at a vacuum or filled wit] 
gas under pressure, such as nitrogen, to resist evaporation of the mercury 
high temperatures. Gas-filled mercury thermometers, as the last form is call 
are so designed that for the whole range of mercury expansion the press 
of the gas opposing it does not rise enough to offer material resistance to 
expansion of mercury or to unduly stress the glass container. It should 
noted that mercury thermometers do not measure the expansion of mere! 
alone, but the difference between the volume of mercury and the glass 'en vekj 
but this is of no consequence so long as this difference is in proportion to 
expansion of the mercury itself, which it is substantially, with proper gi 
composition, when the range is not too great. Such thermometers indid 
temperature changes by the rise and fall of mercury in the stem, and any nun: : 
cal value that may be convenient can be given to any position of the merer 
or any change of position. Common acceptance of certain locations of the s I'j 
number, however, must be recognized as rendering other possible ones unnej* 
sary and so undesirable. Two such scales are recognized, one in use with m 
units, the centigrade, and the other with measurements in English units, 
Fahrenheit, both of which must be known and familiar, because of the freq 
necessity of transformation of numerical values and heat data from one syi= 
to the other. To permit of the making of a scale, at least two points/mus |b 
fixed with a definite number of divisions between them, each called one de t 
The two fixed points are first, the position of the mercury when the thermom 
is in the vapor of boiling pure water at sea level, or under the standard ati 
pheric pressure of 29.92" = 760 mm. of mercury absolute pressure, 
second, the position of the mercury when the thermometer is surrounde 
melting ice at the same pressure. These are equivalent to the boiling- or 



HEAT AND MATTER 



409 



densation, and melting- or freezing-points, of pure water at one atmosphere 
pressure. The two accepted thermometer scales have the following character- 
istics with respect to these fixed points and division between them : 

THERMOMETER SCALES 





Pure Water 
Freezing-point, 
at one atm. pr. 


Pure Water 
Boiling-point, 
at one atm. pr. 


Number of Equal Divisions 
Between Freezing and Boiling. 


Centigrade scale 
Fahrenheit scale 



32 


100 
212 


100 
180 



From this it appears that a degree of temperature change is on the centigrade 
scale, -j-J-Q- of the linear distance between the position of the mercury surface 
at the freezing- and boiling-points of water, and on the Fahrenheit scale. T J-^ of 
the same distance. From this the relation between a degree temperature change 
for the two scales can be given. 



One degree temperature 1 _180_9_ 
change centigrade j 100 5 



of one degree temperature 
change Fahrenheit; 



or 



One degree temperature ) _5 f of one degree temperature 
change Fahrenheit J 9 { change centigrade. 

It is also possible to set down the relation between scale readings, for when the 
temperature is C., it is 32 F., and when it is 100 C. it is (180+32) =212 F., 
so that 

g 
Temperature Fahrenheit = 32 +-^ (Temperature centigrade), 



or 



. Temperature centigrade = ^ (Temperature Fahrenheit -32). 

For convenience of numerical work tables are commonly used to transform 
temperatures from one scale to the other and such a transformation is shown 
in a curve, Fig. 119, and in Table XXIX at end of the Chapter. 

By reason of the lack of absolute proportionality between temperature 
and effect, other fixed points are necessary, especially at high temperatures, 
and the following of Table XVI have been adopted by the U. S. Bureau of Stand- 
ards and are considered correct to within 5 C., at 1200 C. 



410 



ENGINEERING THERMODYNAMICS 



100 i 

JO 
90 -- 

85 - ---? 

75 / 

70 / 
20 

5 i: 

o-- 

66 
60 . 
46 - -- I- 

i- il|| 
f/. 




!=;;[[=!===l===!===!in=:i5!= J;=====!N iitSliiillnHii 

L M Degrees 

S g 2 8 S S 1 I S Fahrenheit 

to . . * 


i i i i i i i i i j 320 1 I 1 1 1 1 1 1 1 ' 
_2- ._ 

: ::z :: sisljJ 

--^-- - 310 -- 
\\-_\\ ~ 305 

- 300 - - - - ^ - 

- 295 -- ::::;; 

- 290 - - - T- - - 

285 H 

u 100 140 i 

::-?: 2*0 :: 
2 I 

/ - 275 -- 






1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 lik I4J 1 1 1 1 1 1 1 1 1 1 1 1 1 fils Htl 1 1 1 1 1 1 1 1 1 1 1 Ilils 1 1 1 1 1 1 1 1 1 LILiy^ 
1 1 1 1 L 1 1 1 1 1 1 1 I s 

^aquaaq^ saaaSaa s; --- "T:"3I qi S ^s 

.-_ _. -- --_ --- - 


tf 860 
10 760 

LO 000 

:ii:ni;2 


: / 

86 -- 

16 -- 
. i 

10 - / 

-20 

-^ 

/ 

-70 y^ 

-70 


10 20 
i i i I i i i i i MI 160 - - 
1 . . . f 
'-'-'- 1~- 155 - - '.'.'.]'. 

- "?-- 160 --- 

\\r\\\ / - 
(-- 145 - - j - - 

60 7 
lit 

130 
130 - - 

m l 

:::z!::: 12( :: :::?: 

-^-- 11.-, Z-- 

y no ]/ 

105^ - 
-30 40 


- 270 - - 
- 265 -- 7 
- 260 -- ----- 
- 265 :;;j;;; 

- 250-^-- 
J 80 120 U 

--?- 246-- 
- -Z 

/ 240 - - 

::::::::: 235:::::::xi 

/ - 230 
-_. .- _-.. 2 - 

- 225 - - - - - Z - - 

. 220 ?--- 

- 216 - 2" 

U 60 100 1 

Degrees Centigrade 


-Z-- - 850 

- 800 - 
- 750 - '.~-~fl- 

- 700 -_/_-- 

3iO I 

40 

i i i i i i [ i i 1 1 650 -- 
- _ _ . / _ 
_-_Z-- 600-- 

/ - 550 /--- 

i/ IN \y\ 
'f-\< | w J^t 
t: o - 

- 420 -- 
370 -/--- 

10 120 100 2 


LO 560 
10 460 
300 
LO 260 



FIG. 119. Granhical Relation between Centierrade and Fahrenheit Thermometer Scales. 



HEAT AND MATTER 



411 



TABLE XVI 
FIXED TEMPERATURES 

U. S. BUREAU OF STANDARDS 



Temperature, 


Temperature, 


Determined by the Point at which 


232 


449 


Liquid tin solidifies 




327 


621 


Liquid lead solidifies 




419.4 


787 


Liquid zinc solidifies 




444.7 


832.5 


Liquid sulphur boils 




630.5 


1167 


Liquid antimony solidifies 




658 


1216 


Liquid aluminum, 97.7% pure, 


solidifies 


1064 


1947 


Solid gold melts 




1084 


1983 


Liquid copper solidifies 




1435 


2615 


Solid nickel melts 




1546 


2815 


Solid palladium melts 




1753 


3187 


Solid platinum melts 





Thermometers in which a liquid and its vapor exist together, depend on a 
operty to be noted in detail later, the relation of vapor pressure to tempera- 
re and its independence of the volume of vapor. So long as any vapor exists 
bove the liquid the temperature will depend only on the pressure of that vapor 
o that such thermometers will indicate temperature by the pressure measure- 
aent, after experimental determination of this pressure-temperature relation 
f vapors. Conversely, temperature measurements of vapors by mercury ther- 
nometers will lead to pressure values, and at the present time some steam 
lants are introducing mercury thermometers on the boilers and pipe lines, in 
lace of the proverbially inaccurate pressure gages. 

Gas thermometer, is the name generally applied to the class in which the 
uid is a gas, whether air, hydrogen, nitrogen or any other, and whether the 
ressure is measured for a fixed contained volume, or the Volume measured 
rhen acted on by a constant pressure. These gas thermometers are so bulky 
s to be practically useless in ordinary engineering work and are only employed 
a standards for comparison and for tests of extraordinary delicacy in investi- 
ation work. They give much larger indications than mercury thermometers 
>ecause the changes of gas volume under constant pressure are far greater 
ban for mercury or any other liquid. Regnault was the first to thoroughly 
avestigate air thermometers and reported that the second form, that of constant 
as volume with measurement of pressure, was most useful. 

Using the centigrade scale, fixing freezing point at C., and making the 
prresponding pressure po, atmospheric at this point, and reading at 100 C. 
mother pressure pioo, he found experimentally a relation between these two 
'pressures and the temperature corresponding to any other pressure p, as 
ven by the empiric formula, 



(601) 



412 ENGINEERING THERMODYNAMICS 

He also determined the pressure at the boiling-point to be related to the pressu] 
at the freezing-point, by 

pioo = 1.3665 po, 
which on substitution gives 



== 272.85 --l ...... (60! 

.3665j>o \po / 

This constant, 272.85, the reciprocal of which is .003665, is, of course, the pressui 
increase factor per degree C. rise of temperature for a gas held at const ai 
volume, received extended investigation and it was found that it had abo. 
the same value applied to the other type of thermometer in which gas volum< 
are measured at constant pressure. This was true even when the pressu] 
used was anything from 44 to 149 cm. of mercury, though it is reporte 
that for 44 cm. pressure the value 272.98, and for 149 cm. pressure, the vain 
272.7, seemed closer. For hydrogen it was found that the constant was sul 
stantially the same as for air, while for carbonic acid it was 270.64, and whi 
the hydrogen thermometers agreed with the air over the whole scale, showir 
proportional effects, this was hardly true of carbonic acid. Such Uncertain! 
in the behavior of these thermometers and in the fixing of the constants w; 
traced to the glass in some cases, but there still remained differences charg 
able only to the gases themselves. Comparison of the air with mercury the 
mometers showed that there was not a proportional change with the temperatu 
and that temperatures on the two, consistently departed. 

Examination of Eq. (602) , giving the relation between two temperatures ar 
the corresponding gas pressures, will show a most important relation. If in E 
(602), the pressure be supposed to drop to zero and it is assumed th 
the relations between pressure and temperature hold, then when p = 
=272.85. This temperature has received the name of the absolute ze 
and may be defined as the temperature at which pressure disappears or becom 
zero at constant volume, and correspondingly, at which the volume also d 
appears, since it was found that similar relations existed between volume ai 
temperature at constant pressure. Calling temperature on a new scale begi 
ning 272.85 below the centigrade zero by the name absolute temperature 
then 

[Absolute temperature 1 f Scale temperature] 

centigrade centigrade J 

.' ' 

As this constant or absolute temperature of the centigrade scale zero, is 
experimental value, it is .quite natural to find other values presented by diff< 
ent investigators, some of them using totally different methods. One of th( 
methods is based on the temperature change of a gas losing pressure withe 
doing work, generally described as the porous plug experiment, and the resu 



HEAT AND MATTER 413 



s the Joule-Thomson effect, and another is based on the coefficient of expansion 
f gases being heated. Some of these results agreed exactly with Regnault's 
alue for hydrogen between C. and 100 C. for which he gave -273 C.= 
-491.4 F. Still other investigations continued down to the last few years 
ielded results that tend to change the value slightly to between 491.6 F., 
nd 491.7 F., and as yet there is no absolute agreement as to the exact value. 
a engineering problems, however, it is seldom desirable or possible to work 
o such degrees of accuracy as to make the uncertainty of the absolute zero a 
.latter of material importance, and for practical purposes the following values 
my be used with sufficient confidence for all but exceptional cases which are 

|i be recognized only by experience. 

. 

[Centigrade = 273) 
Absolute Temperature (T) j Fahrenheit = 460 [ +^ ca ^ e Temperature (Q. 

great accuracy is important it is not possible at present to get a better 
''ahrenheit value than 459.65, the mean of the two known limits of 459.6 and 
59.7, though Marks and Davis in their Steam Tables have adopted 459.64, which 
3 very close to the value of 459.63 adopted by Buckingham in his excellent 
bulletin of U. S. Bureau of Standards and corresponding to 273.13 on the 
tigrade scale. 

These experiments with the gas thermometers, leading to a determination 
f temperature as a function of the pressure change of the gas held at constant 
olume, or its volume change when held at constant pressure, really supply a 
'finition of temperature which before meant no more than an arbitrary number, 
.d furnished a most valuable addition to the generalization of relations between 
at content of a body and its temperature or physical state. 

A lack of proportionality between thermometer indication and temperature, 
as already been pointed out, and it is by reason of this that two identical ther- 
omcters, or as nearly so as can be made, with absolute agreement between 
,ter boiling- and freezing-points, will not agree at all points between, nor will 
e best constructed and calibrated mercury thermometers agree with a similarly 
1 gas thermometer. 

The temperature scale now almost universally adopted as standard is that 
I the constant volume hydrogen gas thermometer, on which the degree F. 
lone one-hundred-and-eightieth part of the change in pressure of a fixed 
blume of hydrogen between melting pure ice, and steam above boiling pure 
Jlter, the initial pressure of the gas at 32 being 100 cm. =39.37 ins. Hg. A 
rcury in glass thermometer indication is, of course, a measure of the proper- 
of the mercury and glass used, and its F. degree of temperature is defined 
parallel with the above as one one-hundred-and-eightieth part of the volume 
the stem between its indications at the same two fixed points. A comparison 
the hydrogen thermometer and two different glasses incorporated in mercury 
rmometers is given below, Table XVII, from the Bulletin of the U. S.. Bureau 
Standards, v. 2, No. 3, by H. C. Dickinson, quoting Mahlke, but it must be 



414 



ENGINEERING THERMODYNAMICS 



remembered that other glasses will give different results ana even differer 
thermometers of the same glass when not similarly treated. 

TABLE XVII 

FAHRENHEIT TEMPERATURES BY HYDROGEN AND MERCURY 
THERMOMETERS 



Temperature by 
Hydrogen 
Thermometer. 


Difference in 
Reading by 
Mercurv in Jena 
59" Glass. 


Difference in 
Reading by 
Mercury in 16" 
Jena Glass. 


Temperature by 
Hydrogen 
Thermometer. 


Difference in 
Reading by 
Mercury in Jena 
59" Glass. 


Difference in 
Reading by 
Mercury in 16' 
Jena Glass. 


32 








617 


+ 10.6 




212 








662 


+ 16.6 




302 




- .18 


707 


+18.7 




392 


+ 1.3 


+ .072 


752 


+24.6 




428 




+ .39 


797 


+28.2 




464 




+ .83 


842 


+38.3 




500 




+1.79 


887 


+41.4 




536 




+2.4 


932 


+50.0 




572 




+3.53 









Useful comparisons of air, hydrogen, nitrogen, and other gases, with alcoh< 
and mercury, in various kinds of glass, are given in the Landolt-Bornsteii 
Meyerhoffer, and in the Smithsonian Physical Tables, but are seldom neede 
for engineering work. 

One sort of correction that is often necessary in mercury thermometi 
work is that for stem immersion. Thermometers are calibrated as a rule wi1| 
the whole stem immersed in the melting ice or the steam, but are ordinari J 
used with part of the stem exposed and not touching the substance whose ter 
perature is indicated. For this condition the following correction is recoi 
mended by the same Bureau of Standards Bulletin : 

Stem correction = .000088 n (t ti) F 

When n = number of degrees exposed; 

t = temperature indicated Fahrenheit degrees; 

ti =mean temperature of emergent stem itself, which must necessarii 

be estimated and most simply by another thermometer next i* 

it, and entirely free from the bath. 

Prob. 1. What will be the centigrade scale and absolute temperatures, for it 
following Fahrenheit readings? -25, 25, 110, 140, 220, 263 scale, and 30(, 
460, 540, 710, 2000 absolute. 

Prob. 2. What will be the Fahrenheit scale and absolute temperatures, for the folk I 
ing centigrade readings? -20, 10, 45, 80, 400, 610 scale, and 200, 410, 65 j, 
810, 2500 absolute. 

Prob. 3. By the addition of a certain amount of heat the temperature of L 
quantity of water was raised 160 F. How many degrees C. was it raised? 



HEAT AND MATTER 415 

Prob. 4. To bring water from C. to its boiling-point under a certain pressure 
required a temperature rise of 150 C. What was the rise in Fahrenheit degrees? 

Prob. 6. For each degree rise Fahrenheit, an iron bar will increase .00000648 of 
its length. How much longer will a bar be at 150 C. than at C.? At 910 C. 
absolute than at 250 C. absolute? 

Prob. 6. The increase in pressure for S0 2 for a rise of 100 C. is given as .3845 at 
constant volume. What would have been absolute zero found by Regnault had he 
used S02 rather than air? 

Prob. 7. A thermometer with a scale from 40 F. to 700 F. is placed in a thermome- 
ter well so that the 200 mark is just visible. The temperature as given by the 
i thermometer is 450. If the surrounding temperature is 100 F., what is true tempera- 
ture in the well? 

4. Calorimetry Based on Proportionality of Heat Effects to Heat Quantity. 

| Units of Heat and Mechanical Equivalent. Though it is generally recognized 

from philosophic investigations extending over many years, that heat is one 

manifestation of energy capable of being transformed into other forms such 

its mechanical work, electricity or molecular arrangement, and derivable from 

them through transformations, measurements of quantities of heat can be made 

without such knowledge, and were made even when heat was regarded as a 

substance. It was early recognized that equivalence of heat effects proved 

effects proportional to quantity; thus, the melting of one pound of ice can cool 

a pound of hot water through a definite range of temperature, and can cool 

wo pounds through half as many degrees, and so on. The condensation of 

i pound of steam can warm a definite weight of water a definite number of 

iegrees, or perform a certain number of pound-degrees heating effect in water. 

3o that taking the pound-degree of water as a basis the ratio of the heat liberated 

iy steam condensation to that absorbed by ice melting can be found. Other 

ubstances such as iron or oil may suffer a certain number of pound degree 

hanges and affect water by another number of pound-degrees. The unit 

>f heat quantity might be taken as that which is liberated by the condensation 

f a pound of steam, that absorbed by the freezing of a pound of water, that to 

fcise a pound of iron any number of degrees or any other quantity of heat 

ffect. The heat unit generally accepted is, in metric measure, the calorie, 

r the amount to raise one kilogramme of pure water one degree centigrade, 

in English units, the British thermal unit, that necessary to raise one pound 

jf water one degree Fahrenheit. Thus, the calorie is the kilogramme degree 

entigrade, and the British thermal unit the pound degree Fahrenheit, and the 

Itter is used in engineering, usually abbreviated to B.T.U. There is also 

occasionally used a sort of cross unit called the centigrade heat unit, which is 

he pound degree centigrade. 

The relation between these is given quantitatively by the conversion table 
t the end of this Chapter, Table XXX. 

All the heat measurements are, therefore, made in terms of equivalent 
rater heating effects in pound degrees, but it must be understood that a water 
ound degree is not quite constant. Careful observation will show that the 



416 ENGINEERING THERMODYNAMICS 

melting of a pound of ice will not cool the same weight of water from 200 F. 
to 180 F., as it will from 60 F. to 40 F., which indicates that the heat capacity 
of water or the B.T.U. per pound-degree is not constant. It is, therefore, 
necessary to further limit the definition of the heat unit, by fixing on some 
water temperature and temperature change, as the standard, in addition to the 
selection of water as the substance, and the pound and degree as units of capacity. 
Here there has not been as good an agreement as is desirable, some using 
4 C. = 39.4 F. as the standard temperature and the range one-half degree 
both sides; this is the point of maximum water density. Others have used 15 
C. = 59 F. as the temperature and the range one-half degree both sides; still 
others, one degree rise from freezing point C. or 32 F. There are good 
reasons, however, for the most common present-day practice which will prob- 
ably become universal, for taking as the range and temperatures, freezing- 
point to boiling-point, and dividing by the number of degrees. The heat unit 
so defined is properly named the mean calorie or mean British thermal unit; 
therefore, 

Mean calorie = (amount of heat to raise 1 Kg. water from C. to 100 C.), 
1UU 

Mean B.T.U.-rgg (amount of heat to raise 1 Ib. water from 32 F. to 212 F.) 



In terms of the heat unit thus defined, the amount of heat per degree tern 
perature change is variable over the scale, but only in work of the most accurat< 
character is this difference observed in engineering calculations, but in accurat 
work this difference must not be neglected and care must be exercised in usinj 
other physical constants in heat units reported by different observers, to be sur 
of the unit they used in reporting them. It is only by experience that judgmen. 
can be cultivated in the selection of values of constants in heat units reporte 
for various standards, or in ignoring differences in standards entirely. Th; 
great bulk of engineering work involves uncertainties greater than these diffei ( 
ences and they may, therefore, be ignored generally. 

By various experimental methods, all scientifically carried out and exetndin 
over sixty years, a measured amount of work has been done and entirely coi 
verted into heat, originally by friction of solids and of liquids, for the dete 
mination of the foot-pounds of work equivalent to one B. T. U., when tl 
conversion is complete, that is, when all the work energy has been converted in 
heat. This thermo-physical constant is the mechanical equivalent of he< 
Later, indirect methods have been employed for its determination by calcu) 
tion from other constants to which it is related. All of these experimer 
have led to large number of values, so that it is not surprising to find doubt 
to the correct value and different values are used even by recognized autho 
ties. The experiments used include: 



HEAT AND MATTER 417 

v 

(a) Compression and expansion of air; Joule. 

(b) Steam engine experiments, comparing heat in supplied and exhausted 

steam; Hirn. 

(c) Expansion and contraction of metals; Edlund and Haga. 

(d) Specific volume of vapor; Perot. 

(e) Boring of metals; Rumford and Him. 
(/) Friction of water; Joule and Rowland. 
(g) Friction of mercury; Joule. 

(h) Friction of metals; Hirn, Puluj, Sahulka. 

(i) Crushing of metals; Hirn. 

(j) Heating of magneto electric currents; Joule. 

(k) Heating of disk between magnetic poles; Violle. 

(I) Flow of liquids (water and mercury) under pressure; Hirn, Bartholi. 

(m) Heat developed by wire of known absolute resistance; Quintus Icilius, 

Weber, Lenz, Joule, Webster, Dieterici. 
(ri) Diminishing the heat contained in a battery when the current produces 

work; Joule, Favre. 
(o) Heat developed in, and voltage of Daniell cells; Weber, Boscha, Favre 

and Silberman, Joule. 
(p) Combination of electrical heating and mechanical action by stirring 

water; Griffiths. 
(q) Physical constants of gases. 

The results of all of these were studied by Rowland in 1880, who himself 
tperimented also, and he concluded that the mechanical equivalent of heat 
fas nearly 

778.6 ft.-lbs. = l B.T.U., at latitude of Baltimore, 



774.5 ft.-lbs. = l B.T.U., at latitude of Manchester. 

fith the following corrections to be added for other latitudes. 

ititude 10 20 30 40 50 60 70 80 90 

ft.-lbs 1.62 1.50 1.15. .62 .15 -.75 -1.41 -1.93 -2.30 -2.43' 

Since that time other determinations have been made by Reynolds and 
.[orby, using mechanical, and Griffiths, Schuste and Gannon, Callendar and 
iarnes, using electrical transformation into heat. Giving these latter deter- 
linations equal weight with those of Joule and Rowland, the average is 

i 

1 small calorie at 20 C. (nitrogen thermometer) =4. 181 XlO 7 ergs. 



418 ENGINEERING THERMODYNAMICS 






On the discussion of these results by Smith, Marks and Davis accept and use 
the mean of the results of Reynolds, and Morby and Barnes, which is 

1 mean calorie = 4. 1834 X10 7 ergs, 

= 3.9683 B.T.U. 
1 mean B.T.U. = 777.52 ft.-lbs., 

when the gravitational constant is 980.665 cm. sec 2 , which corresponds to 32.174 
Ibs., and is the value for latitude between 45 and 46. 

For many years it has been most common to use in engineering calculations 
the round number 778, and for most problems this round number is still th< 
best available figure, but where special accuracy is needed it is likely that nc 
closer value can be relied upon than anything between 777.5 and 777.6 for the 
above latitude. 

' Example. To heat a gallon of water from 60 F. to 200 F. requires the heai 
equivalent of how many foot-pounds? 

1 gallon =8.33 Ibs., 
200 F. -60 F. =140 F. rise, 
8.33 X 140 = 1665 pound-degrees, 
= 1665 B.T.U. 
= 778X1665, ft.-lbs. 
= 90,800 ft. -Ibs. 

Prob. 1. A feed-water heater is heating 5000 gallons of water per hour froi 
40 F. to 200 F. What would be the equivalent energy in horse-power units? 

Prob. 2. A pound of each of the following fuels has the heating values as give: 
Change them to foot-pounds. 

Average bituminous coal, 14,000 B.T.U. per lb., Average kerosene, 18,000 B.T.U. per l| 
" small anthracite, 12,000 B.T.U. per lb., " alcohol, 10,000 B.T.U. per 
Average gasolene, 20,000 B.T.U. per lb. 

Prob. 3. A cubic foot of each of the following gases yields on combusture, t 
number of heat units shown. Change them to foot-pounds. 

Natural gas (average), 880 B.T.U. per cu.ft., Carburetted water gas, 700 B.T.U. per cu. 
Coal gas, 730 B.T.U. per cu.ft., Mond gas, 150 B.T.U. per cu.ft. 

Blast furnace gas, 100 B.T.U. per cu.ft. 

Prob. 4. A pool contains 20,000 cu.ft. of water and must be warmed from 40 
to 70 F. How much work might be done with the equivalent energy? 



HEAT AND MATTER 419 

Prob. 6. How many calories and how many centigrade heat units would be 
required in Prob. 4? 

Prob. 6. In the course of a test a man weighing 200 Ibs. goes up a ladder 25 ft. 
high, every 15 minutes. If the test lasted 12 hours how many B.T.U. did he expend? 

Prob. 7. A reservoir contains 300 billion gallons of water which are heated each 
year from 39 F. to 70 F. What is the number of foot-pounds of work equivalent? 

Prob. 8. A pound of water moving at the rate of 450 ft. per second is brought 
to rest, so that all of its energy is turned into heat. What will be the temperature rise? 

Prob. 9. For driving an automobile 30 horse-power is being used. How long 
will a gallon of average gasolene, sp.gr. = .7, last, if 10% of its energy is converted 
into work? 

Prob. 10. Power is being absorbed by a brake on the flywheel of an engine. 
If the engine is developing 50 horse-power how many B.T.U. per minute must be 
carried off to prevent burning of the brake? 

5. Temperature Change Relation to Amount of Heat, for Solids, Liquids, 
Gases, and Vapors, not Changing State. Specific Heats. Provided gases do not 
decompose, vapors condense, liquids freeze or evaporate, and solids melt, under 
addition or abstraction of heat, there will always be the same sort of relation 
between the quantity of heat gained or lost and the temperature change 
for all, differing only in degree. As the reception of heat in each case 
causes a temperature rise proportional to it and to the weight of the sub- 
stances, this constant of proportionality once determined will give numerical 
relations between any temperature change and the corresponding amount of 
heat. Making the weight of the substance unity, which is equivalent to the 
consideration of one pound of substance, the constant of proportionality may 
be defined as the quantity of heat per degree rise, and as thus defined is the 
specific heat of the substance. Accordingly, the quantity of heat for these cases 
is equal to the product of specific heat, temperature rise and weight of substance 
heated. 

The heat, as already explained, may be added in two characteristic ways: 
(a) at constant volume or density, or (6) at constant pressure. It might be 
expected that by reason of the increase of volume and performance of work 
under constant pressure heating, more heat must be added to raise the tempera- 
ture of one pound, one degree, than in the other case where no such work is done, 
and both experimental and thermodynamic investigations confirm this view. 
There are, therefore, two specific heats for all substances, capable of definition : 

(a) The specific heat at constant volume, and 

(6) The specific heat at constant pressure. 

These two specific heats are quite different both for gases and for vapors, which 
suffer considerable expansion under constant pressure heating, but for solids 
and liquids, which expand very little, the difference is very small and is to be 
neglected here. As a matter of fact, there are no cases of common engineering 
practice involving the specific heat of liquids and solids under constant volume, 
and values for the specific heats of liquids and solids are always without further 
definition to be understood as the constant pressure values. 



420 ENGINEERING THERMODYNAMICS 

Let C, be the specific heat of solids and liquids suffering no change of state. 
" C p , be the specific heat of gases and vapors at constant pressure and 

suffering no change of state. 
" C v , be the specific heat of gases and vapors at constant volume and su 

fering no change of state. 

" t 2 and ti, be the maximum and minimum temperatures for the process. 
" w, be the weight in pounds. 

Then will the heat added, be given by.the following equation, if the tempera- 
ture rise is exactly proportional to the quantity of heat, or in other words, 
if the specific heat is constant. 

Q = Cw(t2 ti), for solids and liquids . V (603) 

Q = C,w(t2 ti), for gases and vapors (not near condensation) when 

volume is constant. . (604) 

Q = C P w(t 2 1\), for gases and vapors (not near condensation) when 

pressure is constant. . . (605) 

When, however, the specific heat is variable, as is the case for many sub- 
stances, probably for all, the above equation cannot be used except when 
the specific heat average value, or mean specific heat is used. If the variatior 
is irregular this can be found only graphically, but for some substances the 
variation is regular and integration will give the mean value. It has been 
the custom to relate the specific heat to the temperature above the freez 
ing-point of water, expressing it as the sum of the value at 32 F., anc 
some fraction of the temperature above this point to the first and seconc 
powers, as in Eq. (606). 

Specific heat at temperature (t) =a+&(*-32)+c(Z-32) 2 . (606 

In this equation a is the specific heat at 32, while b and c are constants 
different for different substances, c being generally zero for liquids. 

When this is true, the heat added is related to the temperature abov-f 
32 by a differential expression which can be integrated between limits 

Q = C* * 2 [a+b(t-32)+C(t-32) 2 ]dt 

Jti -32 

Q9"\3 ft 39^^31 (&(\7 
OA) (li OA) J. ^OU< 



Usually the heats are calculated above 32 so that the heats between an 
two temperatures will be the difference between the heats from 32 to thos 
two temperatures. In this case i = 32, and, t2 = t, whence 



B.T.U. per lb., from 32 to *,= U+|(*-32)+|(*-32) 2 ](*-23). . . 



(608 



HEAT AND MATTER 421 

For this range of temperature 32 to t, the quantity of heat may be ex- 
Dressed as the product of a mean specific heat and the temperature range 
)r 



Heat from 32 to t = (mean sp. heat from 32 to t) X (t 32). . (609) 
Comparing Eq. (608) with Eq. (609) it follows that 

( Mean specific heat 1 ,&/, OON.C/, o 9 x 2 /f. 1n x 

. Q9ir +^ /o ^ ) -a-t-o^-o^-r-^-,^ (oio) 



jfrom 32F.to*F.J ' 2 V " ^3 

The coefficient of (32) in the mean specific heat expression, is half that in 
j,he expression for specific heat at t, and the coefficient of {t 32) 2 , is one-third. 
Fhis makes it easy to change from specific heat at a given temperature 
ibove 32, to the mean specific heat from 32 to the temperature in question. 

The specific heats of some substances are directly measured, but for some 
|)thers, notably the gases, this is too difficult or rather more difficult than cal- 
;ulation of values from other physical constants to which they are related. 
- It often happens that in engineering work the solution of a practical 
problem requires a specific heat for which no value is available, in which case 
-he general law of specific heats, known as the law of Dulong and Petit, for 
ilefinite compounds may be used as given in Eq. (611). 

(Specific heat of solids) X (atomic weight) = 6.4. . . . (611) 

This is equivalent to saying that all atoms have the same capacity for 
seat, and while it is known to be not strictly true, it is a useful relation in 
lie absence of direct determinations. Some values, experimentally determined 
or the specific heats of solids, are given in Table XXXI at the end of this Chap- 
er, together with values calculated from the atomic weights to show the degree 
f agreement. The atomic weights used are those of the International Com- 
aittee on Atomic Weights (Jour. Am. Chem. Soc., 1910). When the specific 
eat of a solid varies with temperature and several determinations are avail- 
ble, only the maximum and minimum are given with the corresponding tem- 
eratures, as these usually suffice for engineering work. 

To illustrate this variability of specific heat of solids, the values deter- 
lined for two samples of iron are given in Figs. 120 and 121, the former 
bowing the variation of the mean specific heat as determined by Oberhoffer 
nd Harker from 500 F. up, and the latter the amount of heat per pound of 
xm at any temperature above the heat content at 500 F., which is gen- 
ially called its total heat above the base temperature, here 500 F. 
; It is extremely probable that the specific heats of liquids all vary irregularly 
tilth temperature so that the constant values given in Table XXXII at the end 
f the Chapter must be used with caution. This is certainly the case for water, 
Id is the cause of the difficulty in fixing the unit of heat, which is best solved 
!y the method of means. In Fig. 122 are shown in curve form the values for the 



422 



ENGINEERING THERMODYNAMICS 






specific heats of water at temperatures from 20 F. to 600 F., as accepted 
by Marks and Davis after a critical study of the experimental results of 

















.18 


































(a) 
(M 


Obert 
Hark 


offer 
r 






/ 


^ 


X 


_ 










(a) 


Jl6 
tt 














// 






sU 


(!>) 








Mean Specific He 














x ^ 


/ 
























x^ 


,' 






















x 


x x 






















.10 






f 




























































1( 


00 





^Temperature in Degrees Fahr. 

FIG. 120. Mean Specific Heat of Iron above 500 F., Illustrating Irregular Variations not 

Yielding to Algebraic Expression. 

Barnes and Dieterici and adjustment of the differences. The integral curve 
is plotted in Fig. 123 which, therefore, gives the heat of water from 32F. to any 























/ 


'(a) 








B.T.Us.-Per Pound of Iron Above 500] 
8 1 1 




(a) 
W 


From 


Oberhof fer D 
Harker 


ata 




s 


/ 






















s 


^ 


\, '. 


















/ s 


^' 


S 




















,S 


<y 






















S 


7* 




















-'" 


s' 

























500 



1000 



2500 



1500 2000 

Temperature in Degrees Fahr . 

FIG 121. Total Heat of Iron above 500 F., Illustrating its Approximation to a Straigl 
Line Relation in Spite of Wide Variation in Specific Heat Given in Fig. 120. 

temperature up to the highest used in steam practice and which is designate 
in steam tables, summarizing all the properties of water and steam, as tl 



HEAT AND MATTER 



423 



100 



200 300 400 

Temperature Degrees Fahr. 



500 



FIG. 122. Specific Heat of Water at Various Temperatures. 



/ 



/ 



200 400 

Temperature in Degrees Fahr. 



600 



IG. 123. Total Heat of Water from 32 F., to any Temperature, the Heat of the Liquid at 

that Temperature above 32 F. 



424 



ENGINEERING THERMODYNAMICS 



eat 



heat of the liquid. For the purpose of comparison, the mean specific h 
of water is given in Fig. 124 from 32 F. to any temperature which is obtained 
from the heat of the liquid above 32 F. by dividing it by the temperature 
above 32 F. 

In the table of specific heats of liquids there is a column giving the value 
calculated from the atomic weights to show at a glance the degree with 
which liquids satisfy the Dulong and Petit law. 

Variability of specific heat is especially noticeable in liquids that are solu- 
tions with different amounts of dissolved substance, in which case the specific 
heat varies with the density and temperature. Problems of refrigeration 
involve four cases of this kind: (a), calcium, and (6), sodium chloride, 



L06 



1.04 



1.0? 



LOO 




200 400 600 

Temperature in Degrees Fahrenheit 

FIG. 124. Mean Specific Heat of Water from 32 to any Temperature. 

brines, the densities of which vary considerably but which are used with 
but little temperature range, seldom over 20 F. and often not over 5 F. ? 
(c), anhydrous ammonia and (d), carbonic acid. 

As the density of brines is often reported on the Baume scale and liquid 
fuels always so, a comparison of this with specific gravities is given in Table 
XXXIII in connection with the specific heat tables at the end of this Chapter 
to facilitate calculation. 

One of the best-known solutions so far as accuracy of direct experimental 
data is concerned, is calcium brine, results for which, from 35 C. to 20 C. 
given below, are from U. S. Bureau of Standards Bulletin by Dickinson, 
Mueller and George, for densities from 1.175 to 1.250. For chemically pure 



HEAT AND MATTER 



425 



calcium chloride in water, it was found that the following relation be- 
tween density D, and specific heat C, at C., 



D = 2.8821 - 3.6272C+ 1.7794C 2 , 



(612) 



and these results plotted in Fig. 125 show the specific heat variation with 
temperature to follow the straight line law very nearly. This being the case 
the mean specific heat for a given temperature range is closely enough the 
arithmetical mean of the specific heat at the two limiting temperatures. To 
fthe figure are also added dotted, the specific heats for some commercial brines, 
not pure calcium chloride, but carrying magnesium and sodium chloride of 
density 1.2. 

It might be conveniently noted here that the relation between freezing- 
point and density for pure calcium chloride by the same bulletin is given 
in Table XVIII below: 

TABLE XVIII 
FREEZING-POINT OF CALCIUM CHLORIDE 

U. S. BUREAU OF STANDARDS 



Density of Solution. 


Per cent CaClt by Wt. 


Freezing-point, 
C. 


Freezing-point, 


1.12 


14.88 


- 9 


15.8 


1.14 


16.97 


-13 


8.6 


1.16 


19.07 


-16 


3.2 


1.18 


21.13 


-20 


- 4.0 


1.20 


23.03 


-24 


-11.2 


1.22 


24.89 


-29 


-20.2 


1.24 


26.77 


-34 


-29.2 


1.26 


- 28.55 


-40 


-40.0 



Other values for the specific heats of brines as commonly used are given 
in Table XIX, the accuracy of which is seriously' in doubt and which 
may be checked by more authoritative values at different points where deter- 
minations have been made. 

Anhydrous ammonia liquid, has a variable specific heat with temperature, 
.put the experimental values are too few to make its value and law quite certain, 
lieveral formulas have been proposed, however, that tend to give an impression 
(of accuracy not warranted by the facts though quite convenient in preparing 
tables. 

Authority Specific heat of NHi liquid at t" F. 

Zeuner 1.0135+.00468 0-32) (a) 

Dieterici 1.118 +.001156 0-32) (6) 

Wood 1.1352+.00438 0-32) (c) 

Ledoux 1.0057 +.00203 0-32) (d) 



(613) 



426 



ENGINEERING THERMODYNAMICS 



I 
.75 

I 



.70 



.GO 




-10 



D=l 



D=l. 



=1.22, 



70 



10 30 50 

Temperature in Degrees Fahr. 

FIG. 125. Specific Heat of Calcium Chloride Brine of Various Densities D at Temperatures 

-10 F. to +70 F. 



HEAT AND MATTER 



427 



TABLE XIX 
SPECIFIC HEAT OF SODIUM CHLORIDE BRINE 



Density, B6 


Sp.gr. 


Per cent NaCl 
by Wt. 


Sp. Heat. 


Temp. F. 


Authority. 


1 


1.007 


1 
1.6 


.992 

978 


-0 
64 4 


Common 
Thomsen 






4 9 


995 


66 115 




5 
10 


1.037 
1.073 


5.0 
10.0 
10.3 


.960 
.892 
.892 


-0 
-0 
59 120 


Common 
Common 
Teudt 






10 3 


912 


59 194 


Teudt 






11.5 


.887 


61-126 


Marianne 






12 3 


871 


64 4 


\Vink elm a,nn 


15 


1.115 


15.0 

18 8 


.892 
841 


-0 
63 125 


Common 
Teudt 






18.8 


.854 


68 - 192 


Teudt 


19 


1.150 


20.0 
24.3 


.829 
.7916 


64-68 


Common 
Winkelmann 






24.5 


.791 


64 


Thomsen 


23 


1.191 


25 


.783 




Common 



From these expressions the mean specific heat follows by halving the coefficient 
of (t 32) F., and these were determined and plotted to scale, together with some 
direct experimental values of Drewes, in Fig. 126. Giving greatest weight 
[to Drewes and Dieterici, a mean curve shown by the solid line is located as 
the best probability of the value for liquid anhydrous and it has the Eq. (614). 



Mean specific heat of anhydrous 1 _i rt7 
liquid NH 3 from 32 F.toW. 



-32). . . . (614) 



From this value the heat of liquid ammonia above 32 F. has been determined 
and is presented graphically in Fig. 127 from which, and the equation, the 
tabular values at the end of the Chapter were determined. 

Ammonia dissolved in water, giving an aqueous solution as used in the absorp- 
tion refrigerating system, has a nearly constant specific heat so closely approxi- 
mating unity as shown by Thomsen, who gives 

3 per cent NH 3 in water solution.^?.^, = .997, at 66 F. 

1.8 per cent NH 3 in water solution, sp.ht., =.999, at 66 F. 

.9 per cent NH 3 in water solution, sp.ht., =.999, at 66 F., 

Ithat it is customary in these calculations to ignore any departure from unity, 

the value for water. 

Liquid carbonic acid, another important substance in engineering, especially 
pn mechanical refrigeration, is less known as to its specific heat than is ammonia, 
land that is much too uncertain. There is probably nothing better available 

at present for the necessary range than the results of Amagat and Mollier, 

reported by Zeuner for the heat of the liquid, which are reproduced in Fig. 

128, and used in the table at the end of this Chapter. 



428 



ENGINEERING THERMODYNAMICS 



It is, however, with gases that the most complex situation exists with respect 
to specific heats. As has already been pointed out, gases may be heated at 



1.4 



I 1 - 2 



OQ 

I" 



50 



-"A 




AA-Zeuner 

BB-Ledoux 

CC-Wood 

DD -Dietetic! 

E.E Drewes 

FF-Mean Used in Book 



+ 50 
Temperature in Degrees Fahr. 



150 



FIG. 126. Mean Specific Heat of Liquid Anhydrous Ammonia from 50 F. to 150 .F 



I 

t3 

S 

i 



wu 
100 



D 

E 


















































/ 
















































/ 


^ 














































/ 


/ 
















































/ 
















































/ 

















































/ 


/ 














































, 


/ 
















































/ 


^ 














































Cx 


, 














































^ 


^ 














































A 
0g 


X^ 






AA- Zeuner 
B- Wood Note: All Curves 
C Dieterici Practically Coin- 
D- Ledoux cident above 32 
E- Mean 
























** 






















A 


^ 


^ 


^ 




























^ 


^ 




























^ 




















































ivi: 50 50 100 150 200 



Temperature in Degrees Fahr. 
FIG. 127. Heat of Liquid Anhydrous Ammonia above 50 F. 

constant volume, doing no external work while being heated, or at constant 
pressure, in which latter case work is done by expansion of the gas against the 
resisting constant pressure. Therefore, there must be two different specific 



HEAT AND MATTER 



429 



heats for each gas, one C v at constant pressure and the other C c at constant 
volume, the difference between them representing the heat equivalent of the 



50 



25 

I 



i_l 

M 
O 

3 



-25 



7 



7 



-25 25 <o 

Temperature in Degrees Fahr. 
FIG. 128. Heat of Liquid Carbonic Acid above 32 F. 

)rk of expansion done during the rise of temperature. Most experimental 
leterminations of the specific heats of gases have been made at constant pressure 



430 ENGINEERING THERMODYNAMICS 

and the constant volume value found from established relations between it 
and other physical constants. These relations most commonly used are two, 
Eq. (615) connecting the difference with a constant R and the other Eq. (616) 



777.52(C P -C,) = R, 

^ = Y (616) 

Op 

/ 

connecting their ratio to a constant y. These constants have each 
a special significance that may be noted here and proved later, thus R is the 
ratio of the PF product of a pound of gas to the absolute temperature, and y the 
particular value taken by the general exponent s in PV s = c, when the expansion 
represented takes place with no heat addition or abstraction, i.e., adiabatic, 
it is also a function of the velocity of sound in gases. Table XXXIV at the end 
of this Chapter gives some authentic values, with those adopted here designated 
by heavy type. 

Variability of specific heats of gases and vapors is most marked and of some 
engineering importance, because so many problems of practice involve highly 
heated gases and vapors, the most common being superheated steam and the 
active gases of combustion in furnaces, gas producers and explosive gas engines, 
In fact, with regard to the latter it may be regarded as quite impossible witrJ 
even a fair degree of accuracy to predict the temperature that will result in thei 
gaseous products from the liberation of a given amount of heat of combustio 
The first fairly creditable results on the variability of the specific heats of gas 
of combustion at high temperatures were announced by Mallard and LeChatelie 
Vieille and Berthelot, all of whom agree that the specific heat rises, but wh 
do not agree as to the amount. A general law was proposed by LeChatelie 
giving the specific heat as a function of temperature by an equation of tl 
following form: 

Specific heat at * F., (F=C),=C, = a+6(*-32), ._ (61 

in which a = specific heat at constant volume at 32 F. This yields, 

jB.T.U.perlb.from32 )_ n ^ 



-32) ..... (61 
J Mean specific heat from 1 , 6 , 



The specific heat at constant pressure is obtained by adding a consta 
to the value for constant volume according to 






whence 

Specific heat at t F., 



HEAT AND MATTER 

~Y\ /"f t\> 



431 



(621) 



-S^. . (622) 



Mean specific heat from 1 _ 
32 F., toZF., (P = C) ~ 



_ r / _ . 
~ Up ~ " t 



6. 



, . 

(b } 



The values of these constants have been determined by LeChatelier, Clerk, 
I Callender, and Holborn and Austin, from which the following values are 
I selected. 

TABLE XX 
SPECIFIC HEAT CONSTANTS, GASES, 



Gas. 


a 


R 

+777T52 


6 


b 

2 


Authority. 


CO 2 
C0 2 

N 2 
N 2 
N 2 


.1477 
.170 


.1944 
.2010 
.2404 
.2350 
2350 


.000097 
.0000824 
.0000484 
.000021 
0000208 


.0000484 
.0000412 
.0000242 
.0000105 
0000104 


LeChatelier 
Holborn and Austin 
LeChatelier 
Holborn and Austin to 2606 F. 
Callender 1544 F to 2440 F 


2 
H 2 O 
1 Air 


.1488 
.3211 


.2125 
.2431 


.0000424 
.000122 
.000135 


.0000212 
.000061 
. 0000675 


LeChatelier 
LeChatelier 
Callender (1544 F. to 2440 F.) 















For purposes of comparison the following curves are plotted, showing all 
these results of specific heat at constant volume, at temperature t F., the total 
heat above 32 F. per pound of gas, and the mean specific heat from 32 F. to 
t F. in Fig. 129. 

Probably there is now more known of the specific heat of superheated steam 
than of any common gaseous substance, and it is likely that other substances 
will be found in time to have somewhat similar characteristics. Pure computa- 
tion from the laws of perfect gases indicates that the specific heat of gases or 
| superheated vapors must be either a constant, or a function of temperature 
s| only, and this is what prompted the form of the LeChatelier formula. Bold 
I experimentation on steam, disregarding the law, or rather appreciating that 
superheated steam is far from a perfect gas, principally by Knobloch and 
f Jacob and by Thomas, showed its specific heat to be a function of both pres- 
sure and temperature. Results were obtained that permitted the direct solu- 
iftion of problems of heat of superheat, or the heat per pound of vapor at any 
[temperature above that at which it was produced, or could exist in contact with 
jjthe liquid from which it came. Critical study of various results by Marks and 
4 Davis led them to adopt the values of Knobloch and Jacob with slight modifi- 

1 



432 



ENGINEERING THERMODYNAMICS 



^ 



\ 



\ 



\r\ 












\ 



\\ 



1 J 

5 03 " 

2 fe 

> S <N 

!o> w 

'1 I 





1 

*2 '~ 

~ % 

I 

o 






\ 




N^N 


s. 




\ . 




\ 


\ 




V 








a 


'i 

. 1 



^ 
> Q- J 

J 1 


Curve 7 Holborn & Austin's Values for COo 
- 8 - " N/ 










^^ 




*\ 


V^ 


\ 




\ 


V 




\ 


















\ 




\ s 


\ 


\ 




\ 


N 




\ 




















N^ 


, s 


\ 


N^ 




\ 




\ 




\ 






Curves 1-6 from values of 
LeChatelier and Clerk 
Solid Lines = Sp.Ht. at Cons 
Broken" 

1 ftn.l 1 POn 9. nnrl S - Vn 
















' 


\ 


\ 


iV 




\ 


\ s 


s 


\ 
























\ 


X 




\ 


y \ 


\ 




\ 






















N 


^ 


>, 






\ 


\ 





























S. N 


* 


\N 






\ 
























S 


\ 


\ 


\ 


\ 


\ 




























\ * 
\ 


^4 N 


\ 




V 


\ 


































^ \ 


\ 






s\ 


































\ 




N 




\ s 


\. 







HEAT AND MATTER 



433 



I 

m 



fl'X'lI.-uoi^jn^s 9AOQB UIROJS jo puno c t aoj ^OH 



oanssojj ^UB^SUOO ^1 IBOH oO'^^tlS a3W 




434 ENGINEERING THERMODYNAMICS 






cations, for which evidence was in existence, raising the specific heats at low 
pressures and temperatures, and their conclusions are adopted in this work. 
In Fig. 130 is shown (A) the Marks and Davis modification of the C p curve 
*of Knobloch and Jacobs, the integral of which (C) gives the heat of superheat 
from any temperature of steam generation to actual steam temperature, while 
(B) shows the values for the mean specific heat above the temperature of satura- 
tion for the particular pressure in question. 

When substances of the same class are mixed so that Wi, W2, w^, etc., Ibs. 
of the different substances having specific heats Ci, C 2) 3, etc., or C P i, C P 2, 
C P 3, etc., or Cci, ,2, C P s, etc., then the specific heat of the mixture is given by 



f . 



Example. If 5 Ibs. of olive oil at a temperature of 100 F., 10 Ibs. of petroleum 
at a temperature of 150 F., and 50 Ibs. of water at 50 F. are mixed together, what 
will be the resultant temperature and how much heat will be required to heat the mix- 
ture 100 above this temperature? 

Sp. ht. of olive oil =* .4, 
Sp. ht. petroleum = .511, 
Sp. ht. water =1.000. 

Let z=the final temp. The heat given up by the substances falling in tempera- 
ture is equal to that gained by those rising, hence 

50(x-50) XI = 5(100 -a;) X. 4 +10(150 -a;) X.511, 
50x -2500 =200 -2x +766 -5.113, 
57.11z =3466, or, y, =60.7 F., 



Sp.ht. of mixture = 1 " /1 ^ c ^" r ^ 3 > f rom Eq (611) 

Wi -\-W 2 +1#3 

5X.4+10X.511+50X1 _57.11 
5+10+50 = 65 

whence the heat required will be 65 X. 8786 = 57 B.T.U. 

Prob. 1. To change a pound of water at 32 F. to steam at 212 F. requires 
1150.4 B.T.U's. If the same amount of heat be given to a cubic foot of each of 
the following substances at 32 F., what will be final temperature in each case? (a) cop- 
per; (6) iron; (c) mercury; (d) clay; (e) stone. 



HEAT AND MATTER 435 

Prob. 2. How many pounds of the following substances could be warmed 10 F. 
y the heat required to raise 100 Ibs. of water from 40 F. to 200 F.? 
(a) Ethyl alcohol from 100 F.; 
(6) Sea water from 60 F., (density = 1.045); 

(c) Glycerine from 60 F; 

(d) Tin from 480 F. 

Prob. 3. If 150 Ibs. of water at 200 F. are added to a tank containing 200 Ibs 
.f petroleum at 70 F., what will be the resultant temperature, neglecting any heat 
Absorbed or given up by the tank itself? 

Prob. 4. To melt 1 Ib. of ice requires 144 B.T.U. How much would this lower 
he temperature of 1 Ib. of the following substances (1) at constant pressure; (2) at 
sonstant volume; (a) air; (6) oxygen; (c) ammonia; (d) hydrogen; (e) nitrogen? 

Prob. 5. What would be the specific heats of the following mixture? Hydrogen 
\ Ibs., oxygen 1 Ib., nitrogen 7 Ibs., carbon dioxide 20 Ibs., carbon monoxide 10 Ibs.? 
Prob. 6. Air is approximately 77 per cent N2, and 23 per cent 02 by weight. By 
paeans of the specific heats of the components, find its specific heats at constant pres- 
jure, and at constant volume. 

Prob. 7. By means of the specific heats, find the values of R and y most correct 
tit atmospheric temperature (60 F.) for, hydrogen, air, carbon dioxide, carbon monoxide 
a,nd nitrogen. 

Prob. 8. How much water could be heated from 40 F. to 60 F. by the heat 
needed to superheat 10 Ibs. of steam at 200 Ibs. per square inch absolute to 700 F.? 
Prob. 9. A building containing 250,000 cu. ft. of space is heated by a hot-water 
:3ystem. Considering the air to change eight times per hour, how many pounds of 
prater per hour must be circulated if the drop in temperature of the water is from 
J200 to 100 and the temperature of the outside air is 30 F. while that of the room 
jis 60 F. neglecting wall conducted heat? 

Prob. 10. How much heat would be required to warm a pound of liquid C0 2 from 
(zero to 80 F.? Compare with water and ammonia. 

6. Volume or Density Variation with Temperature of Solids, Liquids, Gases 
(and Vapors, Not Changing State. Coefficients of Expansion. Coefficients 
j of Pressure Change for Gases and Vapors. Solids increase in length or in any 
1 linear dimension, a certain fraction of their original length for each degree 
temperature rise and the expansion is usually assumed to be in proportion 
to temperature rise. The relation between original and final length can be 
[set down in an equation involving the coefficient of expansion. 

Let a = coefficient of linear expansion = fractional increase in length per 
I degree. 

" h and ti = original length or any other linear dimension and the cor- 
responding temperature; 

and t2 = length which h becomes after heating and the corresponding 
temperature. 

Increase in length = h Ii = ali(t2 ti), (627) 

New length Z 2 = Zi+aZife h), 

fe-*i)] (628) 



436 ENGINEERING THERMODYNAMICS 

Solids, of course, expand cubically and the new volume will be to the old 
as the cubes of the linear dimension. 

Let a = coefficient of volumetric expansion; 
vi = original volume; 
V2 = final volume after heating. 
Then when the temperature rises one degree, 

= l+a .... (629 

If a is small, and it is generally less than ioooo> then the square and cube cai 
be neglected in comparison with the first power, whence 

l+a = l+3a and a = 3a. 

so that the coefficient of volumetric expansion may be taken as sensibb 
equal to three times the coefficient of linear expansion, and similarly, th> 
coefficient of surface expansion as twice the coefficient of linear expansion. 

Liquids, by reason of the fact that they must always be held in solu 
containers, may be said to have no linear expansion, and therefore, althougl 
the expansion may be one direction only, the amount is due to the tota 
change of volume rather than the change of length along the direction o 
freedom to expand. The same is true of gases, so that for gases and liquids 
only coefficients of volumetric expansion are of value and these are given in th 
Tables XXXV, XXXVI, XXXVII, and XXXVIII at the end of this Chapter 
With liquids and gases it is usual to take the volume at C. or 32 F. anc 
29.92 ins. Hg pressure as a standard, and the coefficient gives the increase a:< 
a fraction of this, per degree departure from the freezing-point. This is tht 
universal practice with gases. 

It appears that the coefficients of expansion for solids are quite differen 
from one another, ranging from over 15X10" 4 for wax, to .085X10 ~ 4 fo 
Jena normal glass, a range of over two hundred and sixty times. Determina 
tions of the value at various temperatures for any one substance indicate i 
variation with temperature, which proves that proportionality of increase o 
dimensions to temperature rise, does not hold true, a. fact which has led t< 
formulas of the form 



the value of which is dependent on the determination of the constant and veri 
fication of correctness of form, which has not by any means been conclusively 
done. For most engineering work the constant values nearest the temperature 
range will suffice except for certain liquids, vapors, and gases. A more markec 
tendency to follow such a law of variation with temperature is found witl 
liquids and coefficients for some are given in the standard physical tables. 



HEAT AND MATTER 437 

The two important liquids, mercury and water, have been separately 
studied in greater detail and the latter exhibits a most important exception 
to the rule. For mercury, according to Broch 

which exhibits a refinement of value only in instrument work such as barometers 
ind thermometers. Water, as already mentioned, has its maximum density 
at 39.1 F. and expands with both fall and rise of temperature. Its expansion 
jb given by a similar formula by Scheel, as follows: 

. . (631) 

VTost commonly the expansion of water is not considered in this'way, but by 
omparing densities at varying temperatures, and all sets of physical tables 
ontain values which in this work are significant only as affecting the change 
f volume in turning water to steam and such values as are needed are 
ncorporated in the steam tables later. 

The study of the expansion of gases and vapors at constant pressure, and 
ise of pressure at constant volume, per degree has perhaps been fairly com- 
lete and is of greatest significance, because from it most of the important laws 
f thermodynamics have been derived. This work may be said to have 
tarted with the Regnault air and gas thermometer work, already described, 
ome of the authentic values collected in the Landolt, Bornstein, Myerhoffer, 
ind Smithsonian Physical Tables, are given at the end of this Chapter, 
here a p is the coefficient of pressure change at constant volume, and <x p the 
coefficient of expansion, or volume change at constant pressure. 

The remarkable thing about the coefficients for these gases and vapors is the 
pproach to constancy for most of the gases, not only of the coefficients of expansion 
or P = c nor the similar constancy of the coefficients of pressure rise for V = c, but 
wre remarkable than either of these is the similarity of the two constant coeffi- 
lents. These facts permit of the generalizing of effect when P = c, 
|d when V = c, and of the announcement of a law by means of which 
fl such problems can be solved instead of applying separate coefficients for 
very substance and every different temperature necessary for solids and 
quids where, for example, the maximum coefficient was over 260 times as 
reat as the least. The average coefficient for all gases, applying both to 
pressures and volumes, is the same as enters into the gas thermometer work 
lid its best value is found to be 

a = | = .002034, per degree F. ' 

, . . , . (632) 

a = 27^3 = -003661 , per degree C. 



438 ENGINEERING THERMODYNAMICS 

and approximately 

a = J- = .00203, per degree F. 

a = J- =.00366, per degree C. , 

i o 



. (633) 



These are the same as the reciprocals of the absolute temperature of the 
ice-melting point, and are but expressions of conditions for reduction of the 
volume and pressure at the ice-melting temperature to zero by constanl 
pressure and constant volume abstraction of heat respectively, and bj 
stating the amount of reduction per degree give by implication the numbei 
of degrees for complete reduction. J 

Example. The rails on a stretch of railroad are laid so that they just touch whei 
the temperature is 120 F. How much total space will there be between the rallj 
per mile of track at F.? 

For wrought iron a will be nearly the same for Bessemer steel = .00000648. 

Hence the linear reduction in 5280 ft. for a change of 120 F. will be 

5280 X 120 X .00000648 = 4. 1 f t. 

Prob. 1. A steam pipe is 700 ft. long when cold (60 F.), and is anchored at or 
end. How much will the other end move, if steam at a temperature of 560 F. 
turned into the pipe? 

Prob. 2. A copper sphere is one foot in diameter at 50 F. What must be tl 
diameter of a ring through which it will pass at a temperature of 1000 F.? 

Prob-. 3. A hollow glass sphere is completely filled with mercury at F. Whr 
per cent of the mercury will be forced out if the temperature rises to 300 F.? 

Prob. 4. A room 100 ft.xSO ft.XlO ft. is at a temperature of 40 F. The teni 
perature rises to 70 F. How many cubic feet of air have been forced from the roon 

Prob. 5. The air in a pneumatic tire is at a pressure of 90 Ibs. per square inn 
gage and at a temperature of 50 F. Due to friction of the tire on the ground 
running, the temperat re rises to 110 F. What will be the pressure? 

Prob. 6. A brick lighthouse is approximate y 200 ft. high. Should it be exact 
this at F., what would it be at 100 F.? 

Prob. 7. Show that if a glass tube is rigidly held at each end by brackets attach 
to an iron tank it will break if the tank is warmed. 

Prob. 8. From Eq. (618) find the density of water at 60 F., 100 F., 21 
F., and compare with the values in the steam tables. 

Prob. 9. A drum containing C0 2 gas at a pressure of 250 Ibs. per square in 
gage is raised 100 F. above its original temperature. What will be the new pressui 

7. Pressure, Volume and Temperature Relations for Gases. Perfect a 
Real Gases. Formulating the relations between the pressure change at consts 
volume and the volume change at constant pressure, 

Let P and V be the simultaneous pressure and volume* of gas; 

t be its scale temperature at the same time, F. ; 
" T be its absolute temperature at the same time, F. 



HEAT AND MATTER 439 

Chen at constant volume the pressure reached at condition (a) after heating 
rom 32 F. is given by 






- 



Similarly for another temperature t b , the pressure will be 



Vhence 



492 



1 , 
" 



492 



~ = -^ } for V constant, (634) 

-T& It, 

Similarly 

^ = 1^ for P constant (635) 

V 6 J- ft 

jioth Eqs. (634) and (635) are true, for no gas all the time, but very nearly 
[rue for all, under any range of change, and a hypothetical gas is created for which 
It is exactly true all the time, known as a perfect gas, about which calculations 
ian be made as would be impossible for real gases and yet the results of which 
Ire so close to what would be the result with real gases, as to be good enough 
jor engineering practice. Therefore, with a mental reservation as a guard 
Igainst too great confidence in the work, all real gases will be assumed perfect 
j,nd to follow Eqs. (634) and (635) except when experience shows the results 
lire too far wrong to be useful. 

These laws, known by the names both of Charles and Gay-Lussac, are closely 
jissociated with another also doubly named as Boyle's or Mariotte's and like- 
Use an idealization of experimental observations known to be nearly true for 
I) 11 gases. This is to the effect that so long as temperatures are kept constant 
he pressures of gases vary inversely as their volume, or that, 

=*, and, P Q V a = PbV b = constant, for T constant . (636) 

* b V a 

Study of the PV product, for various gases has revealed a good deal on the 
leneral properties of matter, especially as to the transition from one state to 
mother. This is most clearly shown by curves which may be plotted in two 



440 



ENGINEERING THERMODYNAMICS 




HEAT AND MATTEK 441 

ways. To coordinates of pressure and volume a family of equilateral hyper- 
bolas one for each temperature, would represent the true PV = C or isothermal 
relation and any variation in the constancy of the product would be shown by 
its departure from the hyperbola. Still more clearly, however, will the depart- 
ure appear when the product PV is plotted against pressures, constancy of 
product would require all lines to be straight and inconstancy appear by 
departures from straight lines. To illustrate, the data from Young for car- 
bon dioxide are plotted both ways in Fig. 131, from 32 F. to 496 F., the values 
of PV at 32 and 1 atm. are taken as unity on one scale. It appears that up to 
the temperature of 88 F, known as the critical temperature, each isothermal 
plotted to P and PV coordinates consists of three distinct parts: 

(a) a curved line sloping to the right and upwards; 

(6) a straight line nearly or exactly horizontal; 

(c) a nearly straight line sloping upward rapidly and to the left. 

In this region then the isothermals are discontinuous, and this is caused by 
the liquification or condensation of the gas, during which increase of pressure, 
produces no change of volume, provided the temperature is low enough. It 
also appears that each PV line has a minimum point and these minima joined 
result in a parabola. At the end of this Chapter are given in Table XXXIX the 
values of PV at three different temperatures and various pressures for oxygen, 
hydrogen, carbon dioxide and ammonia, in terms of the values at 32 and 1 
atm. for further comparison and use. Further study along these lines is not 
profitable here and the topic while extremely interesting must be dropped with 
the observation, that except near the point of condensation or liquefaction, 
gases or vapors, which are the same thing except as to nearness to the critical 
state, follow the Boyle law closely enough for engineering purposes. 

None of these approximate laws, Eqs. (634), (635) and (636) can be con- 
sidered as general, because each assumes one of the variables to be constant, but a 
general law inclusive of both of these follows from further investigation of a 
fixed mass of gas suffering all sorts of pressure volume and temperature changes, 
such as occur in the cylinders of compressors and gas engines. A table of 
simultaneous experimental values of pressure, volume, and temperature, for any 
gas will reveal the still more general relation inclusive of the preceding three as 
follows : 

P.V. P>V b _PV_ , 

Iv "IT T " 

iii which C g is approximately constant for any one gas and assumed constant 
for perfect gases in all calculations. For twice the weight of gas at the same 
pressures and temperatures C ff would be twice as large, so that taking a constant 
R for one pound, and generally known as the " gas constant," and introducing 
a weight factor w, the general characteristic equation for the perfect gas is, 

(638) 



442 



ENGINEERING THERMODYNAMICS 



This general law may be derived from the three primary laws by imagining 
in Fig. 132, two points, A and B, in any position and representing any two states 
of the gas. Such points can always be joined by three lines, one constant 



Diagram to derive Law ^=C^ 
FIG. 132. Curve of Continuous Relation between P, V, and T for Gases. 

pressure A to X, one constant temperature X to Y, and the other constant 
volume Y to B. For these the following relations hold, passing from A to B 

T 
V V 

V a V xm 



But 



and 



whence 



P 
V V v. 

V X V y-p , 



P T 

V T7 v a 
v a V vr> 7fT> 



or 



Passing to B, 



?sZs = Vz 
T. T,- 



T, 
'TV 



HEAT AND MATTER 443 

But 



or in general 



T 

= fM 

Ty Ty 



PV 

T = constant. = wR 



when the weight of gas is w Ibs. 

For numerical work, the values of R must be fixed experimentally by direct 
measurement of simultaneous pressure, volume, and temperature, of a known 
weight of gas or computed from other constants through established relations. 
One such relation already mentioned but not proved is 

R = 777.52(C P -C V ). . ....... (639) 

It is extremely unlikely that the values of R found in both ways by a multi- 
tude of observers under all sorts of conditions should agree, and they do not, 
but it is necessary for computation work that a reasonable consistency be attained 
and that judgment in use be cultivated in applying inconsistent data. In the 
latter connection the general rule is to use that value which was determined by 
measurement of quantities most closely related to the one being dealt with. 
Thus, if R is to be used to find the state of a gas as to pressure, volume, and tem- 
perature, that value of R determined from the first method should be selected, but 
the second when specific heats or Joule's equivalent are involved. Of course, a 
consistency could be incorporated for a perfect gas, but engineers deal with real 
gases and must be on guard against false results obtained by too many hypoth- 
eses or generalizations contrary to the facts. Accordingly, two values of R are 
given in Table XL, at the end of this chapter, one obtained from measure- 
ments of specific heats at constant pressure and determinations of the ratio 
of specific heats unfortunately not always at the same temperature and gen- 
erally by different people, and the other by direct measure of gas volume at 
standard 32 F. temperature and 1 atm. pressure. These measurements are 
separately reported in Sections (5) and (8), respectively. 

If a gas in condition A, Fig. 133, expand in any way to condition B, then 
it has been shown that 



in which /has any value and which becomes numerically fixed only when the 
process and substance are more definitely defined. Comparing the temperatures 
at any two points A and B, it follows that 



444 
whence 

But 

and 

whence 



ENGINEERING THERMODYNAMICS 



T a PaVa 



PaVa \V 



~ 



and 



T a \V b J 






A 














\ 














\ 














\ 


\ 














X 


\ 


^-^ 


B 


















V 

don or Compression of Gas between A and B, Causing a Change c 
perature. 

Vb /7V\rH 
V~~ V 7\ / 



(640) 



(641) 



Eqs. (640) and (641) give the relation between temperatures and volume? 
But 



= 
V, P a T b 

which, substituted in above, gives 



T b P 



or 



HEAT AND MATTER 445 

and 



o \-- a 

:>r 



irHir , (642) 

J. a \JTn / 



(643) 



Eqs. (629) and (630), give the relation between pressures and temperatures. 
It is convenient to set down the volume and pressure relations again to 
complete the set of three pairs of most important gas equations. 



(644) 
(645) 



iese are perfectly general for any expansion or compression of any gas, but 
we of value in calculations only when s is fixed either by the gas itself or by 

|;he thermal process as will be seen later. 



Example. A pound of air has a volume of 7.064 cu.ft. at a pressure of two atmos-. 
|)heres and a temperature of 100 F. Find the value of R for air from the data; 
,lso the final volume and temperature if expansion occurs so that s=*1.4 until the 
ressure becomes ^ an atmosphere. 

PV=wRT, or 2116x2x7.064 = 1X^X560, or #=53.38, 

s-l .4 



.-. !F 2 = 7^1.49=-- =352 abs. = -108 F. 
1.49 

i 
|? = ^V =2.7, or 7t-2.77i-19.lcu.ft. 

I Prob. 1. A perfect gas is heated in such a way that the pressure is held constant. 
r the original volume was 10 cu.ft, and the temperature rose from 100 F. to 400 
j., what was the new volume? t , 

I Prob. 2. The above gas was under a pressure of 100 Ibs. per square inch gage at 
|je beginning of the heating. If the volume had been held constant what would have 
iaen the pressure rise? 
| Prob. 3. A quantity of air, 5 Ibs. in weight, was found to have a volume of 50 cu.ft. 

id a temperature of 60 F. What was the pressure? 

5 Prob. 4. A cylinder holding 12 cu.ft. has a pressure of 250 Ibs. per square 

||h gage, and the temperature is 50 F. What would be the weight of its contents 

-ere it filled with (a) C0 2 ; (6) NH 3 ; (c) Oxygen; (d) Hydrogen? 



446 ENGINEERING THERMODYNAMICS 

Prob. 5. At a pressure of 14.696 Ibs. per square inch and a temperature of 
melting ice, one pound of air has a volume of 12.387 cu.ft. From the data find the 
value of R for air. The specific heats of air are given by one authority as C p = .2375 
and C P = .1685. Find R from the data and see how the two values obtained compare. 

Prob. 6. At the same pressure and temperature as in Prob. 5, a pound of the 
following substances has a volume as shown. From the data and the values of 
specific heats, find R by the two methods. 

Substance. Cu.ft. per Ib. C p . C 9 . 

Hydrogen 178.93 3.409 2.412 

Carbon dioxide.... 8.15 .217 .1535 

Oxygen 11.21 .2175 .1551 

Nitrogen 12 . 77 . 2438 . 1727 

Prob. 7. 5 cu.ft. of gas at a pressure of 3 atmospheres absolute and a tempera- 
ture of 50 F. expand to atmospheric pressure. What will be the final volume and 
temperature, if s = 1.35? 

Prob. 8. 1000 cu.ft. of gas at atmospheric pressure and 60 F. are compressec 
into a tank of 100 cu.ft. capacity. What will be the pressure in the tank and the 
temperature of the gas at the end of the process, if the gas is C0 2 and the com- 
pression adiabatic? 

Prob. 9. What will be the final volume, pressure and temperature, if a pound o: 
air at atmospheric pressure (14.7 Ibs. per square inch) and a temperature of 60 F 
be compressed adiabatically until its absolute temperature is six times its origina 
value? 

8. Gas Density and Specific Volume and its Relation to Molecular Weigh 
and Gas Constant. The density of a gas is best stated for engineering 
purposes as the weight of a cubic foot, but as this becomes less on rise o 
temperature or decrease of pressure it is necessary to fix a standard condition 
for reporting this important physical constant. It is best to take one atmosphere 
760 mm. or 29.92 ins. of mercury as the pressure, and C. = 32F. as the 
standard temperature, though it is in some places customary in dealing with 
commercial gases, such for example as those used for illumination, to take th 
temperature at 60 F. and illuminating gas at this condition is often knowr 
among gas men as standard gas. In this work, however, the freezing-poin 
and standard atmosphere will be understood where not specifically mentioned 
as the conditions for reporting gas density and its reciprocal, the specific volum 
of gases or the cubic feet per pound. The chart, Fig. 134, shows the relatioi 
of volume and density at any pressure and temperature to the volume am 
density under standard -conditions. 

These constants have been pretty accurately determined by many investi 
gators, whose figures, to be sure, do not agree absolutely, as is always th 
case in experimental work, but the disagreement is found only in the las 
significant figures. Some selected values of reliable origin are reported a 
the end of this Chapter in Table XLI for the important gases and 
may be used in computation work. 



f 



HEAT AND MATTER 



447 



It often happens in dealing with gases and especially superheated vapors 
that a value is needed for which no determination is available, so that general 



Pressure in Pounds Per Sq. In. Abs. 
16 15 14 13 12 



10 




40 ] 50 00 

Temperature , Degrees Fahr. 



,80 



I I 



I I I I 
1.2 1.1 



I ^P 5 I 
i i 



I I 1-56 | 



I I I I I I I 
9 .8 



I I 
.7 



Upper Sea* =Ratlo 
Inner Sea,e . 



Ou t e, Sca,e - 



I I I I I I I I i 

.6 
Volume at 32 29.92" 

Density at 32 F. 
Density at any T 



FIG. 134. Equivalent Gas Densities At Different Pressures and Temperatures. 
, of density or specific volumes of substances are necessary to permit the 
led constant to be estimated. These relations may be applied to vapors 



448 ENGINEERING THERMODYNAMICS 

as well as to gases even though the standard conditions are those for the 
liquid state, on the assumption that all gases and vapors will expand under 
temperature, or contract under pressure rise, to the same degree, retaining 
the same relative relations between all substances as exist at the standard 
atmosphere and freezing-point. A vapor thus reported below its point of 
condensation and assumed to have reached that condition from one of higher 
temperature at which it exists as vapor is often called steam gas, or alcohol 
gas, for example in the case of water and alcohol. 

Such general relations between the densities of gases as are so desirable 
and useful in practical work have been found by studying the manner in which 
gases chemically combine with respect to the volume relations before and after 
the reaction. Following several experimenters, who reported observed rela- 
tions, Gay-Lussac stated a general law, as follows: 

When two or more gaseous substances combine to form a compound, the vol 
umes of the combining gases bear a simple ratio to each other and also to 
that of the compound when it is also a gas. 

He also attempted to derive some relation between this law and Dalton's atomic 
combining law, which states that, in combining chemically, a simple numerica 
relation exists between the number of atoms of different elements which unite 
to form a compound. This was not successful, but Avagadro later found tht 
expected relation by assuming that it is a particle, or a number of atoms, o 
a molecule, that is important in combining, and the law stated is as follows : 

Equal volumes of different gases measured at the same pressure and tempera 
ture contain the same number of molecules. 

It is possible by analysis of these two laws to get a relation between the volume 
of gases and the weights of their molecules because the molecular relation o 
Avagadro, combines with the combining law of Gay-Lussac to define the rela. 
tion between the number of combining molecules. At the same time the weigh 
relations in chemical reactions, based on atomic weights, may be put into 
similar molecular form, since the weight of any one substance entering is th 
product of the number of its molecules present and the weight of the molecuk 
Applying the relation between the number of molecules derived previouslj 
there is fixed a significance for the weight of the molecule which for simple gasc 
like hydrogen and oxygen is twice the atomic weight and for compound gasei 
like methane and carbon dioxide, is equal to the atomic weight. Applyin 
this to the Avagadro law, the weights of equal volumes of different gases mus 
be proportional to their molecular weights, as equal volumes of all contain tl 
same number of molecules. 

Putting this in symbolic form and comparing any gas with hydrogen, as 
its density, because it is the lightest gas of all and has well determined chara! 



HEAT AND MATTER 449 

teristics, requires the following symbols, denoting hydrogen values by the 
subscript H. 

Let m = molecular weight of a gas, 



8 = density in Ibs. per cu.ft. = , 



then 
and 



<> 



But as the molecular weight of hydrogen is for engineering purposes equal to 2 
closely enough and hydrogen weighs .00562 Ib. per cu.ft. = S#, at 32 F and 
29.92 ins. Hg, 

Lbs. per cu.ft. = Si = .00281 mi ...... (648) 

To permit of evaluation of Eq. (648) it is necessary that there be available 
a table of molecular weights of gases and the atomic weights of elements from 
which they are derived, and the values given at the end of this Chapter in Table 
XLII are derived from the international table. As atomic weights are 
purely relative they may be worked out on the basis of any one as unity, and 
originally chemists used hydrogen as unity, but for good reasons that are of no 
importance here, the custom has changed to Y& the value for oxygen as unity. 
These atomic weights are not whole numbers but nearly so, therefore, for con- 
venience and sufficient accuracy the nearest whole number will be used in 
this work and hydrogen be taken as unity except where experience shows it 
to be undesirable. 

The reciprocal expression to Eq. (648) can be set down, giving the specific 
volume of a gas or its cubic feet per pound at 32 F. and 29.92 ins. Hg., as 
follows : 



Cu.ft. per ib.--an- ...... (649) 



This is a most important and useful conclusion as applied to gases and vapors 
[for which no better values are available, and in words it may be stated as follows: 

The cubic feet per pound of any gas or vapor at 32 and 29.92 ins. Hg, is 

equal to 355.87 divided by its molecular weight, 
or 

The molecular weight of any gas or vapor in pounds, will occupy a volume of 
355.87 cu.ft. at 32 and 29.92 ins. Hg. 

The approach to truth of these general laws is measured by the values 
[given for specific volume and density at the end of this Chapter (a) experiment- 
ially derived and, (6) as derived from the hydrogen value by the law. 



450 ENGINEEEING THERMODYNAMICS 

Another and very useful relation of a similar nature is derivable from what 
has been established, connecting the gas constant R with molecular weights. 

ID 

The general law PV = wRT when put in the density form by making ^ = TF 
becomes 

^ = RT. . (650) 



Whence, comparing gases with each other and w^th hydrogen at the same 
pressure and temperature 

Pi 



RH i * 

and Si 



which indicate that the densities of gases are inversely proportional to the 
gas constants, or the density of any gas is equal to the density of hydrogen times 
the gas constant for hydrogen divided by its own. 

Inserting the values of density at 32 and 29.92 ins. Hg and of the gas con- 
stant for hydrogen, it follows that for any gas 

Lbs.percu.ft^S^ ~ f . . ..... (653) 

the reciprocal of which gives the specific volume at 32 F. and 29.92 ins. Hg, or 1 

P 

, (654) 



Example. 1. Explanation and use of Chart, Fig. 134. This diagram is for the- 
purpose of finding the cubic feet, per pound, or pounds per cubic foot, of a gas at32 c 
F. and a pressure of 29.92 ins. of Hg, if its volume or weight per cubic foot be known f 
at any pressure and temperature. The curves depend upon the fact that the pounds 
per cubic foot (8) vary directly as the pressure and inversely as the temperature. 
That is 

, T 29.92 



The line of least slope is so drawn that for any temperature on the horizontal scale 
its value when divided by 492 may be read on the vertical scale. The group of line,' 
with the greater slope is so drawn that for any value on the vertical scale this quantit; 

29 92 
times ~ may be used on the horizontal scale. That is, the vertical scale gives th 



HEAT AND MATTER 451 

itio of densities as affected by temperature for constant pressure, while horizontal 
3ale gives the ratio as affected by both temperature and pressure. A reciprocal 
3ale is given in each case for volume calculations. 

To find the pounds per cubic foot of gas at 32 F. and 29.92 ins. of mercury when 
,s value is known for 90 and 13 Ibs. per sq.in. On the temperature scale, pass 
ertically until the temperature line is reached, then horizontally until the curve 
;>r 13 Ibs. absolute is reached. The value on the scale below is found to be 1.265, 
) that the density under the standard conditions is 1.265 of the value under known 
onditions. Had it been required to find the cubic feet per pound the process would be 
precisely the same, the value being taken from the lower scale, which for the example 
|3ads .79, or, the cubic feet per pound under standard conditions is 79 per cent of 
he value under conditions assumed. 

Example 2. By means of the molecular weight find the density of nitrogen at 
2 F. and 29.92 ins. Hg, and the cubic feet per pound for these conditions. 
From Eq. (646) 

81 mi 28 X. 00562 
g^= , or 81 = . 

lence 8 for nitrogen = .07868 pounds per cu.ft. and, 



Prob. 1. Taking the density of air from the table, find the value of R for air, by 
neans of Eq. (653) and compare its value with that found in Section 7. 

Prob. 2. Compare the density of carbon monoxide when referred to 32 F. and 
SO F. as the standard temperature, as found both ways. 

Prob. 3. By means of their molecular weights find the density of oxygen, nitrogen 
md carbon dioxide at 32 F. and 29.92 ins. Hg. 

Prob. 4. What are the cubic feet per pound of acetylene, methane and ammonia 
t 32 F. and 29.92 ins. Hg? 

Prob. 5. An authority gives the following values for R. Compare the densities 
ound by this means with the densities for the same substance found by the use of 
he molecular weights. 

Oxygen 48.1 

Hydrogen 764.0 

Carbon monoxide 55.0 

Prob. 6. What will be the volume and density under standard conditions, of a 
;as which contains 12 cu.ft. per pound at a temperature of 70 F. and a pressure of 
o Ibs. per square inch absolute? 

Prob. 7. What will be the difference in volume and density of a gas when con- 
jidered at 60 and 29.92 ins. of Hg, and at 32 F. and 29.92 ins. of Hg? 

p 9. Pressure and Temperature Relations for Vapor of Liquids or Solids, 
j'/aporization, Sublimation and Fusion Curves. Boiling- and Freezing-points 

or Pure Liquids and Dilute Solutions. Saturated and Superheated Vapors. 
e jfobstances may exist in one of three states, solid, liquid or gas, the latter being 

generally called vapor when, at ordinary temperatures the common state is that 



452 ENGINEERING THERMODYNAMICS 

of liquid or solid, or when the substance examined is near the point of lique- 
faction or condensation, and just which state shall prevail at any time depends 
on thermal conditions. Within the same space the substance may exist in two oi 
these three states or even all three at the same time under certain special condi 
tions. These conditions may be such as to gradually or rapidly make that parl 
in one state, turn in to another state, or may be such as to maintain the relative 
amounts of the substance in each state constant; conditions of the latter sort are 
known as conditions of equilibrium. These are experimental conclusions, but 
as in other cases they have been concentrated into general laws of which they 
are but special cases. The study of the conditions of equilibrium, whether o 
physical state or chemical constitution, is the principal function of physica 
chemistry, in the pursuit of which the Gibbs phase rule is a controlling prin 
ciple. According to this rule each possible state is called a phase, and th 
number of variables that determine which phase shall prevail or how man> 
phases may exist at the same time in equilibrium for one chemical substanc 
like water, is given by the following relation, which is but one of the conclusion 
of this general principle of equilibrium. 

Number of undefined variables = 3 number of phases. 

Now it is experimentally known that if water be introduced into a vacuun 
chamber some of it will evaporate to vapor and that, therefore, water and it 
vapor may coexist or the number of phases is two, but this does not state 
or when equilibrium will be attained. The rule above, however, indicates tha 
for this case there can be but one undefined or independent variable and, o 
course, since the pressure rises more when the temperature is high than whei 
low, the two variables are pressure and temperature, of which accordingly o 
one is free or independent, so that fixing one fixes the other. In other word 
when a vapor and its liquid are together the former will condense or the latte 
evaporate until either pressure or temperature is fixed, and fixing tiie one th 
other cannot change, so that the conditions of equilibrium are indicated 
a curve to coordinates P and T, on one side of which is the vapor state anc 
on the other that of liquid. Such a curve is the vapor pressure-temperature curv 
of the substance, sometimes called its vapor tension curve, and much experi 
mental information exists on this physical property of substances, all obtaine* 
by direct measurement of simultaneous pressures and temperatures of a vapo 
above its liquid, carefully controlled so that the pressure or the temperature i 
at any time uniform throughout. 

The conditions of equilibrium between vapor and liquid, defined by the vapo 
tension curve extend for each substance over a considerable range of pressur 
and temperature, but not indefinitely, nor is the range the same for each. A 
the high-pressure and temperature end a peculiar interruption takes place du 
to the expansive effect of the temperature on the liquid and the compressiv 
effect of the pressure on the vapor, the former making liquid less dense and tl 
latter making vapor more dense, the two densities become equal at son: 
pressure and temperature. The point at which this occurs is the " critical point 
at which the equilibrium between liquid and vapor that previously existec 



HEAT AND MATTER 



453 



nds and there is no longer any difference between vapor and liquid. This 
k)int is a most important one in any discussion of the properties of matter, 
,nd while difficult to exactly locate, has received much experimental attention, 
nd some of the best values are given below in Table XXI for the pressure, 
lensity, and temperature denning it, for the substances important in engineering' 



TABLE XXI 
THE CRITICAL POINT 



Substance. 


Symbol. 


Critical Temp. 


Critical Pres- 
sures. 


Critical 
Density 
Water 
at 
4C=1. 


Authority. 


Criti- 
cal vol. 
Cu.ft. 
perLb. 


Authority. 


C. 


F. 


Atm. 


Lbs. 
per 
Sq.in. 


lydrogen 
>xygen 

Nitrogen 


H 2 
2 

N 2 
NH 3 

NH; 

C0 2 
C0 2 

H 2 
H 2 
H 2 O 

H 2 
H 2 
H 2 


-243.5 
-118.1 

-146.1 

+130.0 
+131.0 

+ 31.35 
+ 30.921 

+358.1 
+364.3 
+365.0 

+374. 
+374.6 
+374.5 


-390.1 
-180.4 

-232.8 

266. 
267.8 

88.43 
87.67 

676.4 
687.7 
689. 

705.2 
706.3 
706.1 


20 
50i 

35.i 

115. 
113. 

72.9 

77.i 

194.61 
200.5 


294 
735 

515 

1690 
1660 

1070 
1130 

2859 
2944 

3200 
3200 


.652 

.442 

.464 
.452 

.429 


Olszewski 
i Wroblewski 
2 Dewar 
1 Olszewski 
2 Wroblewski 
Dewar 
Vincent and 
Chappuis 
Amagat 
i Andrews 
2 Cailletet and 
Matbiaa 
Nadejdini 
Batteli 
Cailletet and 
Colardeau 
Traube and 
Teichner 
Holborn and 
Baumann 
Marks 


26.8 
13. 


Nadejdini 
Batteli _. 


Linmonia 
iinmonia 


/arbon dioxide. . . 
Carbon dioxide. . . 

Vater 


Vater 


Vater 


Vater 
Pater 




Vater 







To illustrate this discussion there is presented the vapor tension curves of 
ater, ammonia and carbon dibxide to a large scale in chart form derived 
rom the tabular data both at the end of this Chapter, while a small scale dia- 
ram for water is given in Fig. 135. These data are partly direct experimental 
eterminations and partly corrections obtained by passing a smooth 
urve representing an empiric equation of relation between pressure and tem- 
erature, through the major part of the more reliable experimental points. 

se pressure-temperature points are very accurately located for ivater, the 
rst good determinations having been made by Regnault in 1862 and the last 
y Holborn and Baumann of the German Bureau of Standards in the last year, 
lie data presented are those of Regnault corrected by various investigations 
>y means of curve plotting, and empiric equations by Wiebe, Thiessen and 
>chule, and those of various later observers, including Battelli, Holborn, Hen- 
ing, Baumann, Ramsay and Ypung, Cailletet and Colardeau, somes eparately, 
but all together as unified by Marks and Davis in their most excellent steam 



454 



ENGINEERING THERMODYNAMICS 



tables, and later by Marks alone for the highest temperatures 400 F. to the 
critical point, which he accepts as being located at 706.1 F. and 3200 Ibs. square 




Temperature in Degrees Fahr. 
FIG. 135. Vapor of Water, Pressure-temperature Curve over Liquid (Water). 

inch. In calculations the values of Marks and Davis, and Marks, will 
accepted and used. 



HEAT AND MATTER 



455 



Carbon dioxide and ammonia are by no means as well known as steam, 

and the original data plotted, while representing the best values obtainable, must 

be accepted with some uncertainty. A smooth curve Figs. (139) and (140) 

has been drawn for each through the points at locations that seem most fair, 

for both these substances and the values obtained from it are to be used in 

alculations; these curves have been located by the same method as used by 

^Earks in his recent paper and described herein later. The equalized values 

,re given in the separate table at the end of the Chapter with others for latent 



.1 

c 

HH 

0< 
GO 

*H 

G> 

&j 

02 

3.05 

.s 

02 

o> 

i 



i 

-< 










































/ 


















/ 




















/ 
















Ice 




/ 


















/ 


















/ 


A 


ipor 
















/ 
















S* 


/ 















-- 


***?, 


_ 












-35 -10 +15 +40 



Temperature in Degrees Fahr. 
FIG. 136. Vapor of Water, Pressure-temperature Curve over Solid (Ice). 

heats and volumes, but while consistent each with the other are probably but 
Ilittle more correct than values reported by others which are inconsistent. 

The curves. and the equivalent tabular data are most useful in practical 
(work, as they indicate the temperature at which the vapor exists for a given 
i pressure, either as formed during evaporation or as disappearing during con- 
idensation, or the other way round, they indicate the pressure which must be 
maintained to evaporate or condense at a given temperature. 

Just as the vapor-liquid curves indicate the conditions of equilibrium between 



456 



ENGINEERING THERMODYNAMICS 



vapor and its liquid, dividing the two states and fixing the transition pressure 
or temperature from one to the other, so also does a similar situation exist with 
respect to the vapor-solid relations. In this case the curve is that of " sub- 
limation " and indicates the pressure that will be developed above the solid 
by direct vaporization at a given temperature in a closed chamber. In Fig. 
136 is plotted a curve of sublimation of vapor-ice, based on Juhlin's data, 
Table XXII, which indicates that the line divides the states of ice from that 
of vapor so that at a constant pressure, decrease of temperature will cause 
vapor to pass directly to ice and at constant temperature a lowering of pres- 
sure will cause ice to pass directly to vapor. 

TABLE XXII 
JUHLIN'S DATA ON VAPOR PRESSURE OF ICE 



Temperature. 


Pressure. 


C. 


F. 


Min. Hg. 


Lbs. sq.in. 


-50 


-58. 


.050 


.001 


-40 


-40. 


.121 


.0023 


-30 


-22. 


.312 


.006 


-20 


- 4. 


.806 


.0156 


-15 


5. 


1.279 


.0247 


-10 


14. 


1.999 


.0386 


- 8 


17.6 


2.379 


.0459 


- 6 


21.2 


2.821 


.0544 


- 4 


24.8 


3.334 


.0643 


- 2 


28.4 


3.925 


.0758 


- 


32. 


4.602 


.0888 



Likewise the liquid, water, may pass to the solid, ice, by lowering temperature 
at a fixed pressure as indicated in Fig. 137, plotted from data by Tamman, 
Table XXIII, and which becomes then the curve of "fusion." 

TABLE XXIII 

TAMMAN'S DATA ON FUSION PRESSURE AND 
TEMPERATURE OF WATER-ICE 



Temperature. 


Pressure. 


C. 


F. 


Kg. sq.cm. 


Lbs. sq.in. 





32. 


1 


1423 


- 2.5 


27.5 


336 


4779. 


- 5. 


23. 


615 


8747.4 


- 7.5 


18.5 


890 


13658.8 


-10.0 


14. 


1155 


16428. 


-12.5 


9.5 


1410 


20055. 


-15. 


5. 


1625 


23113. 


-17.5 


.5 


1835 


26100. 


-20. 


- 4. 


2042 


27044. 


-22.1 


- 7.8 


2200 


31291. 



HEAT AND MATTER 



457 

























\ 




















\ 


V 




















\ 


s 




















\ 






















\ 




















\ 


\ 




















\ 






















V 


ater 


















\ 


\ 




















\ 














Ice 






\ 


\ 




















I 






















\ 




















\ 






















\ 




















\ 



Temperature in Degrees Fahrenheit 
FIG. 137. Water Ice, Pressure-temperature Curve. 



458 



ENGINEERING THEEMODYNAMICS 






These three curves plotted to the same scale meet at a point located at a pressure 
of 4.6 mm. Hg. = .18 ins. Hg., and temperature +.0076 C. =32.01 F., ordi- 
narily taken at 32 F., which point is named the triple point, as indicated in 
Fig. 138. The fact that the vapor pressure for water extends below freezing- 
point and parallels more or less that of ice indicates the condition of supercooled 



02 



!.l 



Ice 



Water 



-Tri 




Vapor 



)le Point 



50 

Temperature Degrees Fahr. 

FIG. 138. Water Vapor Water Ice, Combined Curves of Pressure-temperature Rela- 
tion. The Triple Print. 

water, one of unstable equilibrium instantly dispelled by the introduction of 
a little ice at the proper stable state for this temperature. 

Ordinary engineering work is not concerned with the entire range indicated 
in Fig. 138 for any substance, but with the higher temperature ranges for some 
and the low for others, with transition from solid to liquid state for metals 
and similar solids and the transition from liquid to vapor for a great many, of 
which water comes first in importance, then the refrigerating fluids, ammonia 



HEAT AND MATTER 459 

and carbon dioxide, and last certain fuels like alcohol and the petroleum oils 
with their distillates and derivatives. 

Melting-points, or the fusion temperature of such solids as are important, 
are usually given for only one pressure, the standard atmosphere, as in ordinary 
practice these substances are melted only at atmosphere pressure, and some 
such values are given at the end of the Chapter in Table XLIII. 

This is not the case, however, for boiling-points, which must be denned 
a little more closely before discussion. The vapor pressure curves indicate 
that as the temperature of a liquid rises, the pressure rises also if the substance 
is enclosed, but if the pressure were relieved by opening the chamber to a region 
of lower pressure and kept constant, then the temperature would no longer 
rise and boiling or ebullition would take place. The boiling-point then is the 
highest temperature to which the liquid and its vapor could rise under the 
existing pressure. When not otherwise defined the term boiling-point must 
be taken to mean the temperature of ebullition for atmospheric pressure of 
29.92 ins. Hg, and values for several substances are given at the end of this 
Chapter in Table XLIV. 

Vapor having the temperature required by the pressure of the pressure- 
temperature curve is known as saturated vapor, and this may be denned as 
vapor having the lowest temperature at which it could exist as vapor, under 
the given pressure. Vapors may, however, be superheated, that is, have 
higher temperatures than saturated vapors at the same pressure, but cannot 
so exist for long in the presence of liquid. Superheating of vapors, therefore, 
implies isolation from the liquid, and the amount of superheat is the number of 
degrees excess of temperature possessed by the vapor over the saturation 
temperature for the pressure. In steam power plant work, especially with 
turbines, it is now customary to use steam with from 75 F. to 150 F. of 
superheat, and it might be noted that all so-called gases like oxygen and 
nitrogen are but superheated vapors with a great amount of superheat. 

It has already been mentioned that the saturated vapor pressure-temperature 
curve of direct experiment is seldom accurate as found, but must be corrected 
by empiric equations or smooth average curves, and many investigators have 
sought algebraic expressions for them. These equations are quite useful also 
in another way, since they permit of more exact evaluation of the rate of change 
of pressure with temperature, which in the form of a differential coefficient 
is found to be a factor in other physical constants. One of these formulas 
for steam as adopted by Marks and Davis in the calculation of their tables 
is given in Eq. (655), the form of which, was suggested by Thiessen: 



0+459.6) lo gl -^ = 5.409(^-212 )-3.71XlO-i[(689-0 4 -477*], . (655) 

in which t = temperature F.; and p = pressure Ibs. sq.in. 

This represents the truth to within a small fraction of one per cent up to 400 
F., but having been found inaccurate above that point Professor Marks has 



460 ENGINEERING THERMODYNAMICS 






very recently developed a new one, based on Holborn and Baumann's 
high temperature measurements, which fits the entire range, its agreement 
with the new data being one-tenth of 1 per cent, and with the old below 400 
F., about one-fifth of 1 per cent, maximum mean error. It appears to be the 
best ever found and in developing it the methods of the physical chemists have 
been followed, according to which a pressure is expressed as a fraction of the 
critical pressure and a temperature a fraction of the critical temperature. 
This gives a relation between reduced pressures and temperatures and makes use 
of the principle of corresponding states according to which bodies having the same 
reduced pressure and temperature, or existing at the same fraction of their 
critical are said to be in equivalent states. The new 'Marks formula is given 
in Eqs. (656) and (657), the former containing symbols for the critical 

I pressure p c I and the latter g j vmg to them their numerical values, 

( temperature T c abs. J 
in pressure pounds per square inch, and temperature absolute F. 

log 2i = 3.006854 (y-l) [l + . 0505476^-+. 629547 (^--.7875 VI, . (656) 

log p = 10.515354 - 4873.71 T- * - .004050967 7 + .000001 392964 T 72 . . (657) 

As the method used in arriving at this formula is so rational and scientific, 
it has been adopted for a new determination, from old data to be sure, of the 
relations between p and T for ammonia and carbon dioxide, so important as 
substances in refrigeration, especially the former. According to this method 
if PC and T c are the critical pressures and temperatures, both absolute, and 
p and T those corresponding to any other point, then according to Van der 
Waals, 

/ m \ 

(658) 



Accordingly, the logarithm of the critical divided by any other pressure, is 
to be plotted against the quantity [(critical temperatur edivided by the tem- 
perature corresponding to the pressure) !], and the form of curve permits of 
the determination of the function, after which the values of the critical 
point are inserted. This has been done for NHs and C02 with the result for 
NH 3 

.... (659) 



which on inserting the critical constants, 
p c =114 atm. = 1675.8 Ibs. per square inch 
T e = 727.4 F. absolute 



which are the Vincent and Chap- 
puis values, 



HEAT AND MATTER 461 

becomes, 

logp = 5.60422-1527.54r~ 1 -171961T~ 2 .... (660) 
For CO2 it was found that 



(661) 

which on inserting the critical constants, 

p c = 77 atm. = 1131.9 Ibs. per square inch 

which are Andrews' values, 
T 7 ^ 547.27 F. abs. 

becomes, 
logp = 7.46581-4405.7657 7 ~ 1 + 1617501.3667 7 ~ 2 -257086165.87067 7 ~ 3 . (662) 

Curves showing the relation of reduced and actual temperatures and pressures 
are given in Fig. 139 for ammonia and in Fig. 140 for carbon doixide. 

For the past half century far more time and effort have been devoted to 
making other formulas of relation of p to T for saturated vapor not only for 
steam, but also for other vapors, than would have sufficed for accurate exper- 
imental determination, and as these help not at all they are omitted here. Equa- 
tions of physical relations can be no better than the data on which they are based, 
and for the substances ammonia and carbonic acid the charts or formulas must 
be used with a good deal of suspicion. 

In all engineering calculations requiring one of these constants even for steam 
no one is justified in using a formula; the nearest tabular or chart value must 
be employed and it will be as accurate as the work requires. Time is at 
least as important as accuracy, if not more so, for if too much time is required 
to make a calculation in commercial work, it will not be made because of the 
cost, indirect and approximate methods being substituted. 

It is sometimes useful in checking the boiling-point of some substance 
little known, to employ a relation between boiling-points of different substances 
at the same pressure applied to a substance well-known. 

Let T a and TD be absolute temperatures of boiling for substances A and B under 

same pressure; 
Td and T b ' be absolute temperature of boiling for substances A and B under 

some other pressure. 
'Then, 

^ = ^+c(7Y-n) ........ . . (663) 

lb 1* 

Such equations as this are useful in finding the saturation curve of other sub- 
L stances from that for water, which is now so well established, when enough 
Ipoints are known for the other substances to establish the constant c. Also 

T ' 
the ratio -~ plotted against the temperature difference T b ' T b should give 



462 



ENGINEERING THERMODYNAMICS 





















































^ 


Logarithm of (Critical Pressure -f- any other Pressure) 
Vo ^ os bo o io 4^ as 






































x 


X 






































x 


^ 






































/ 






































s 


X 






































// 


/ 






































X 


X 






































X 






































? 


X 






































y> 






































Jf 


V 








From Values of Wood o 
" " " Dieterici 
" Regnault + 
" " Ledoux / 


















X s 


s> 
























X 


^ 






































X 






































V 


' 






































X 





















































































.2 .3 .4 .5 

(Critical Temperature Divided by any other Temperature)! 



,0 



sure in Pounds Per SqJn.Abs^ ^ 


















































































/ 










































/ 








































1 








































, 


/ 








































/ 








































/ 








































., 


? 








































/ 








































y 








































x 




















CO ^V" 

1 

200 






















x 




From Values of 


Wood o 
Dieterici 
Reg-nault + 
Ledoux / 






















y 


X 






















^ 


Lc^ 



































-^r- 


.r- - 


-^ 

































-40 



10 



200 



60 110 160 210 

Temperatures in Degrees Fahr. 

FIG. 139. Ammonia Pressure-temperature Relations, for Saturated Vapor. 



HEAT AND MATTER 



463 

























































- 


OJ 










































/ 




1 5 








































/ 













































/ 








Logarithm ol (.Critical Pressure^-^- any other 

'en 




































/ 










































^ 










































/ 








































. 


/< 








































/ 


s 








































/ 


m 








































x 








































x* 


S 








































^ 








































x 


^ 






From Cailletet's Data 
u Regnault's ^ 
" Stewart's Interpolation of Zeu 
" Zeuner's Tabulation of Mollier 
based on Amagat's Data o 
















^ 










tier's Data + 
8 Formula 









. 


^ 














^ 


8 








































2 













































.1 .2 .3 .4 .5 

(Critical Temperature Divided by any other Temperature)-! 











































/ 












































/ 






yUU 

* 






































/ 










































/ 








% 


































j 

















From Cailletet's Data 
" Regnault's A 
" Stewart's Interpolation of Zeuner's Data + 
* Zeuner's Tabulation of Mollier's Formula 
based on Amagat's Data o 






/ 










& 
,, Aft 








/ 












m GOO 








/ 














Pressure in Pounds 




























x 


^ 








































k 








































/ 








































X 








































/ 






































/? 


X 






































/ 




































j** 


>^" 


^ 






























_ 


*- 


.- r 







































100 



50 

Temperatures in Degrees Fahr. 



464 



ENGINEERING THERMODYNAMICS 



a straight line, and if the line is not straight the experimental values may b 
wrong or the law untrue. This procedure has been followed in Fig. 141, ii 
checking the curves for CO2 and NH 3 against those for water, but it is impos 
sible to say whether the discrepancies for C02 are due to a failure of the law o 
bad experimental values, probably both, as the law holds poorly for water itself. 



250 



200 



150 



100 



.6 

Values of 



T A ' 
"V 



FIG. 141. Curves for CO 2 and NH 3 to Check the Linear Relation Eq. (663). 

All of the preceding refers, of course, to pure substances, but in practi 
work there are frequently encountered problems on solutions where lai 
differences may exist compared to the pure liquids. Thus, for salts in water 
is well known that addition of a salt lowers the freezing-point, that more s 
lowers it more, and it was first thought that the depression was in proporti 
to the amount dissolved. This being found to be untrue, recourse was h 



HEAT AND MATTER 



465 



again to molecular relations by Raoult, who announced the general law that 
the molecular depression of the freezing-point is a constant. 



Molecular lowering of freezing-point E' = = const., 



(664) 



in which 



A = depression of freezing-point in degrees F. ; 
w = weight dissolved in 100 parts of solvent; 
m = molecular weight of substance dissolved. 

From Eq. .(664) the freezing-point for brines may be found as follows: 
Freezing-point of aqueous solutions = 32 (const.) X . . 



(665) 



As examples of the degree of constancy of the " constant " the following values 
Table XXIV, taken from Smithsonian Tables are given: 

TABLE XXIV 
LOWERING OF FREEZING POINTS 





g. Mol. 


Molecular 




Salt. 


1000 g. H 2 O' 


Lowering. 


Authorities. 


NaCl 


.004 


3.7 






.01 


3.67 






.022 


3.55 


Jones 




.049 


3.51 


Loomis 




.108 


3.48 


Abegg 




.232 


3.42 


Roozeboon 




.429 


3.37 






.7 


3.43 




NH 4 C1 


.01 


3.6 






.02 


3.56 






.035 
.1 


3.5 
3.43 


Loomis 




.2 


3.4 






.4 


3.39 




CaCb 


.01 


5.1 






.05 


4.85 






.1 


4.79 






.508 


5.33 






.946 


5.3 


Arrhenius 




2.432 


8.2 


Jones-Getman 




3.469 


11.5 


Jones-Chambers 




3.829 


14.4 


Loomis 




.048 


5.2 


Roozeboon 




.153 


4.91 






.331 


5.15 




1 


.612 


5.47 






.788 


6 34 





466 ENGINEEKING THERMODYNAMICS 

Just as the~pressure of dissolved substances in liquids lowers the freezing- 
point, so also does it lower the vapor pressure at a given temperature or raise 
the boiling-point at a given pressure. Investigation shows that a similar 
formula expresses the general relation: 

Molecular rise of boiling-point = E = -^ = constant = 5.2, . (666) 

when water is the solvent. 

From Eq. (666) the rise of the boiling-point is found to be 

m 
Rise of boiling-point = 5.2 (667) 

When liquids are mixed, such as is the case with all fuel oils and with 
denatured alcohol, the situation is different than with salts in solution, and 
these cases fall into two separate classes: (a) liquids infinitely miscible like 
alcohol and water or like the various distillates of petroleum with each other, 
and (6) those not miscible, like gasolene and water. 

The vapor pressure for miscible liquid mixtures is a function of the pressure 
of each separately and of the molecular per cent of one in the other when there 
are two. This rule, which can be symbolized, is no use in engineering work, 
because in those cases where such mixtures must be dealt with there will be 
generally more than two liquids, the vapor pressure characteristic and molec- 
ular per cent of each, or at least some of which will be unknown. 

When, however, the two liquids in contact or in fact any number are 
non-miscible they behave in a very simple manner with respect to each other, 
in fact are quite independent in action. Each liquid will evaporate until its 
own vapor pressure is established for the temperature, as if the other were not 
there, and the vapor pressure for the mixture will be the sum of all the separate 
ones. On the other hand the boiling-point will be the temperature at which 
all the vapor pressures together make up the pressure of say the atmosphere, 
and this is necessarily lower than the highest and may be lower than the 
lowest value for a single constituent. This action plays a part in vaporizers 
and carburettors using alcohol and petroleum products. To permit of some 
approximations, however, a few vapor tension curves for hydrocarbons and alco- 
hols are given later in the Section on vapor-gas mixtures, and data on the vapor 
pressure and temperature relations of ammonia-water solution are given in the 
section on the solution of gases in liquids. 

Example 1. Through how many degrees has ammonia vapor at a pressure of 
50 Ibs. per square inch absolute been superheated, when it is at the temperature at 
which steam is formed under a pressure of 100 Ibs. per square inch absolute? 

From the curve of pressure and temperature of steam the temperature is 328 F. 
for the pressure of 100 Ibs. From the similar curve for ammonia vaporization occurs 
under a pressure of 50 Ibs. at a temperature of 22 F. Hence, superheat = 328 
22=306F. 



HEAT AND MATTER 467 

Prob. 1. Three tanks contain the following liquids together: water, ammonia, 
find carbon dioxide respectively, and at a temperature of 30 F. What pressure 
iBxists in each tank? If the temperature rises to 70 F. how much will the pressure 
;:ise in each? 

Prob. 2. The pressure exerted by water vapor in the atmosphere when saturated, 

that due to the temperature and is independent of the pressure of the air. The 
itotal pressure read by a barometer is the sum of the air pressure and the water vapor 
! pressure. What is the pressure due to each under a saturated condition for tem- 
peratures of 50 F., 100 F., 150 F., and 200 F., the barometer in each case being 
29.92 inches of Hg? 

Prob. 3. In order to secure a sufficiently high rate of heat transfer the steam in 
fa, radiator must be at a much higher temperature than the room to be warmed. If it is to 
be 150 above room temperature what must be its pressure for room temperatures 
f 50 F., 60 F., 70 F., 80 F., and 125 F.? 

Prob. 4. In one type of ice machine ammonia gas is condensed at a high pressure 
d evaporated at a low pressure. What is the least pressure at which gas may be 
ndensed with cooling water of 70 F., and what is the highest pressure which may 
be carried in the evaporating coils to maintain a temperature in them of F.? 

Prob. 5. Should carbon dioxide be substituted in the above machine what pressures 
would there be in the condensing coils, and in the evaporating coils? 

Prob. 6. How many degrees of superheat have the vapors of water, ammonia and 
barbon dioxide at a pressure of 15 atmospheres and a temperature of 500 F.? 

Prob. 7. Change the following pressures in pounds per square inch absolute \o 
reduced pressures for water, ammonia, and carbon dioxide, 15 Ibs., 50 Ibs., 100 Ibs., 
500 Ibs. 

Prob. 8. At the temperature of melting ice what will be the vapor pressure of ammonia 
IJnd carbon dioxide? At the temperature of melting tin what will be the pressure of 
{water vapor? At this same temperature how many degrees of superheat would 
ammonia vapor under 100 Ibs. pressure have, and how many degrees superheat would 
carbon dioxide vapor have under 1000 Ibs. pressure? 

Prob. 9. If 10 Ibs. of common salt, NaCl, be dissolved in 100 Ibs. of water, what 
will be the boiling point of the solution at atmospheric pressure, what the freezing-point? 

10. Change of State with Amount of Heat at Constant Temperature. Latent 
Heats of Fusion and Vaporization. Total Heats of Vapors. Relation of Spe- 
cific Volume of Liquid and of Vapor to the Latent Heat. As previously explained, 
i liquid boils or is converted into a vapor at constant temperature when the 
pressure on the surface is constant. Then during the change of state the amount of 
Iheat added is indirect proportion to the amount of vapor formed. The amount of 
vapor to convert a pound of liquid into vapor at any one steady tempera- 
ture, is the latent heat of vaporization some values for which are given at the 
end of this chapter in Table XLV, and it must be understood that this 
natent heat is also the amount given up by the condensation of a pound of vapor. 
I Latent heat is not the same for different pressures or temperatures of vapori- 
sation but is intimately associated with the volume change in the transition 
jifrom the liquid to the vapor state. That this should be so, is clear on purely 
Itational grounds because there is necessarily external mechanical work done 



468 ENGINEERING THERMODYNAMICS 

in converting the liquid to the vapor, since this is accompanied by a change 
of volume against the resisting pressure at which the conversion takes place. 
Thus, if 

Vv = specific volume of the vapor in cubic feet per pound; 
V L = specific volume of the liquid in cubic feet per pound ; 
P = pressure of vaporization Ibs. per sq.ft. absolute. 

Then 

J Mechanical external work done dur- 1 _p/y 17 ) ft lb (66* 

{ ing vaporization of 1 Ib. 

Of course, at high temperatures the volume of a pound of liquid is great* 
than at low because of its expansion with temperature rise, and under the co 
responding higher pressures the volume of a pound of vapor is less, becaus< 
of the compressional effect of the pressure, than at low pressures, so that a 
pressures and temperatures rise the difference VvV L becomes less and dis- 
appears at the critical point where it is zero. The latent heat being thus asso 
ciated with a factor that becomes less in the higher ranges of temperature 
and pressure may be expected, likewise to become less unless some other facto 
tends to increase. All the energy of vaporization making up the latent hea 
may be said to be used up in (a) doing external work as above, or (6) overcom 
ing attraction of the molecules for each other. As at the critical point theri 
is no molecular change and no external work, the latent heat becomes zer< 
at this point. 

This relation between latent heat and volume change was formulated b; 
Clausius and Clapeyron, but Eq. (669) is generally known as Clapeyron' 
equation : 

Let L = latent heat ; 

J = mechanical equivalent of heat = 778, or better 777.52, in sucl 

cases as this; 
T = absolute temperature of vaporization; 

(i dP , . 

-j~i = rate of increase of vapor pressure per degree change of corre 

spending temperature. 
Then 



This formula is used to calculate latent heat from the specific volumes of vapo 
and liquid and from the curvature of the saturation curve when they are known 
but as these volumes are especially difficult to measure, direct experimenta 
determination of the latent heat should be depended upon to get numerica 
values wherever possible. The formula will then be useful for the invers- 
process of calculating specific volumes from latent heats or as a means' o 



HEAT AND MATTER 



469 



checking experimental values of both, one against the other. It is, however, 

iust as useful to calculate latent heats from the specific volumes, and - - of 

&T 

the vapor curve, when the latent heats are less positively determined than 
jhe volumes or densities. 

Another simpler relation of a similar general character exists and is useful 
in estimating latent heats approximately for some little known substances 
ike, for example, the liquid fuels, and in the use of which accurate physical 
iata are badly needed. Despretz announced that 



s nearly constant for all substances, and this was simplified by Ramsay and 
Crouton on the assumption, first, that the volume of the liquid is very small 
it ordinary temperatures and may be neglected, in comparison with the volume 
pf the vapor, and second, that the volume of the vapor is inversely proportional 
p the molecular weight m and directly proportinal to absolute temperatures 
to that (Trouton's law) 



>r 



-m constant = C 



T 
= c 

m 



(670) 



,he constant c is given the following values by Young: 

CO 2 c = 21.3 

NH 3 c = 23.6 

Hydrocarbons c = 20.21 

Water and alcohols c = 26 

For such substances as water and steam, the properties of which must be 
bccurately known, general laws like the above are of no value compared with, 
lirect experimental determination except as checks on its results, and even 
hese checks are less accurate than others that are known. 

These experimental data are quite numerous for water, but as generally 
aade include the heat of liquid water from some lower temperature to the 
oiling-point. The amount of heat necessary to warm a pound of liquid from 
emperature 32 F. to some boiling-point, and to there convert it entirely into 
apor is designated as the total heat of the dry saturated vapor above the origi- 
al temperature. This is, of course, also equal to the heat given up by the con- 
ensation of a pound of dry saturated vapor at its temperature of existence and 
f the subsequent cooling of the water to some base temperature taken univer- 
Mly now as 32 F. in engineering calculations. 



470 ENGINEERING THERMODYNAMICS 

From observations by Regnault and formulated by him in 1863 the present 
knowledge of the total heat of water may be said to date. He gave the 
expression, Eq. (671), in which the first term is the latent heat at 32 and 
one atmospheric pressure: 

Total heat per pound dry saturated steam = H = 1091. 7 +.305 (J-32). (671) 

This was long used as the basis of steam calculations, but is now to be discarded 
in the light of more recent experimental data, the best of it based on indirect 
measurements by Grindley, Griessmann, Peake, who observed the behavior of 
steam issuing from an orifice, together with the results of Knobloch and Jacob 
and Thomas on specific heats of superheated steam, and in addition on direct 
measurements by Dieterici, Smith, Griffiths, Henning, Joly. All this work| 
has been recently reviewed and analyzed by Davis, who accepts 1150.3J 
B.T.U. as the most probable value of the total heat under the standard atmos-| 
phere and the following formula as representing total heats from 212 up tol 
400 F. 

# = 1150.3+.3745(*-212) -. 0005500 -212) 2 . . . (675 

The Davis curve containing all the important experimental points and th 
accepted line, extended dotted from 212 to 32, is presented in Fig. 142. 

From the total heats given by this formula the latent heat is obtainab] 
by subtraction, according to the relation, 

Latent heat (L)= total heat of vapor above 32 F. (J^)- heat of 

liquid from 32 F. to boiling point (h), (673 

in which the heat of the liquid is computed from a mean curve between Dieterici' 
and Regnault's values, having the equation ft = .9983 .0000288 032) -f 
.0002133(<-32) 2 . This is the basis of the values for latent and total heats in th 
Marks and Davis steam tables referred to, and accepted as the best obtain 
able to-day. From these tables a pair of charts for latent heat and total heat 
dry saturated steam are given at the end of this Chapter. 

The specific volume and density of dry saturated steam, given in the chart 
and table are calculated, as this seems to promise more exact results than direc 
experiment, the method of calculation involving three steps: 

(a) From the pressure-temperature equation the ratio of is found b. 
differentiation as follows: 

log p = 10.515354-4873.7ir~ 1 -.00405096T+. 000001392964 7 12 , 
whence 

^f = (-^P^ -.00405096+ .000002785928 T\ p. 



HEAT AND MATTER 



471 



(6) From the latent heats the difference between specific volume of vapor 
jind liquid, (V V -V L ) is calculated by substituting (a) in Clapeyron's equation. 

(c) From the Landolt, Bornstein, Myerhoffer tables for density of water 
bhe volume V L is taken, whence by addition the volume of the vapor V v is found 



For ammonia and carbonic acid there are no data available on total heats 
by either direct measure or by the orifice expansion properties, and very few 



-1083,5/s- 



D 

1069; 



-1112,4;. 



] 126^8 -H, 



-1117-.C 



I/ 



^-1123.2 



-1155 



1IJ7. 



i:jy. 



-1144.8 




1170 



/ 



llWsi 



1159,5 



-1184; 



oX^ 



X 




1166. 



1173, 




32 50 fcS 86 104 122 110 158 176 194 212 230 248 266 284 



320 338 356 374 392 



Temperature in. Degrees K. 
FIG. 142. Total Heat of Dry Saturated Steam above 32 F. (Davis). 

leterminations of the latent heat itself, so that the process that has proved so 
iatisfactory with steam cannot be directly followed with these substances. 
Vccordingly, a process of adjustment has been used, working from both ends, 
Beginning with the pressure temperature relations on the one hand and specific 
volumes of liquid and vapor on the other, the latent heat is determined by 
plapeyron's equation and where this does not agree with authentic values an 
idjustment of both latent heat and specific volume is made. 



472 ENGINEERING THERMODYNAMICS 

This process is materially assisted by the so-called Cailletet and Mathij 
law of mean diameter of the curves of density of liquid and vapor, which 
given in Figs. 143, 144 and 145, for water, ammonia and carbon dioxide, on 
which the points are marked to indicate the source of information. 

On each of these curves the line BD is the line of mean density, its abscissa 
being given by the following general equation, 

(674) 



Of course, this mean density line passes through the critical volume B. For 
these three cases this Eq. (674) is found to have the form, 



(675; 



For water s = 28.7 -.015(- 300) -.000015(*-300) 2 . (a) 

For ammonia s = 20 -.022(^-30). (b) 

For carbonic acid.. s = 33.1 -. 0219(^+20) -.00016(Z-20) 2 . (c) 



A more exact equation for water has been determined by Marks and Davis 
in their steam tables and is 

s = 28.424-.01650(*-320)-.0000132(-320) 2 . , . . (676) 

From the smooth curve, which has the above equation, the volumes and densi- 
ties of liquid and vapor that are accepted have been derived, and are presented 
in chart form on a large scale and in tabular form at the end of the Chapter, 
the values for water being those of Marks and Davis. 

From these volume differences and the relation the latent heats have 

been calculated and the newly calculated points are compared with experimental 
values in Fig. 146. 

The total heats are obtained by adding to the latent heat the heat of 
liquid above 32 from 50 F. up to the critical point for C02 and to 
150 F. for NHs, which include the working range for refrigeration. 
These liquid heats have already been determined in Section 5 in discussing 
specific heats. 

Charts and tables at the end of this Chapter give the final values of total heat, 
heat of liquid, latent heat, specific volume and density of dry saturated 
vapor based on large-scale plottings, without equations beyond that for the 
pressure-temperature relations for saturated vapor, and the results are be- 
lieved to be as reliable as it is possible to get them without more experimental 
data. 

The properties of dry-saturated steam are given in Table XLVII, and' 
charts, A, B, C, D, E, F; the properties of superheated steam, in Table XLVIII; 
dry-saturated ammonia vapor in Table XLIX, and Charts G, H, I, J, K, L; 
and dry-saturated carbon dioxide vapor in Table L, and -Charts M, N, 
O, P, Q, R. 



HEAT AND MATTER 



473 



I 



Solid curves from data of 

Marks and Davis 
Dotted extension to reach 
new critical temperature 
as found by Marks 




400 



Upper Scale =Lbs.per Cu.Ft. 



.025 .02 

Lower Scale =Cu.Ft.per Lb. 



Temperature in Degrees Pabr. 

i 






^ 

\ 














































































V 


\ 








































\ 








































\ 










































\ 








































\ 










































\ 










































\ 








































\ 


s 








































\ 


s. 








































>^ 


X 










































\ 


^ 










































>> 


^ 












































^ 















































*^, 


b -*-^ 







Volume of Vapor in Cu.FU 
FIG. 143. Specific Volume and Density of Liquid and Dry Saturated Vapor of Water. 



ENGINEERING THERMODYNAMICS 



250 

J 

i 

5 

f 

#50 



50' 
( 

150 

1 

? 

j. 

1 50 





-50 

( 

FIG. 


"S- 




1 
















^J 








T 


Gtitic 


\l-Poi 


.^ 

























































s* 














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*, 


^, 


















































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N 












































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<fe 
































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, 




















































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fl 






























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XL 


























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) 5 10 15 20 25 30 ' 35 40 
.2 .1 .075 .05 .04 .03 .025 

Upper Scale= Lbs. per Cu. Ft. Lower Scale =: Cu. Ft. per Lb. 


H 


















































































1 


















































































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t 


















































































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) 5 10 15 gO 
Volume of Yapor.'in Gubio Feet per Ppuad^. 

144. Specific Volume and Density of Liquid and Dry Saturated Vapor of Amrxioni; 



HEAT AND MATTER 



475 



o From Data of Amagat 

" " Cailletet & Mathais 
- Behn 




3 .1 .05 

Upper Scale = Lbs.pen Cu.Ft. 



Lower Scale =Cu.Ft.per Lb. 



80 



Q 40 



pera 
S 



Critical Poll 






Volume of Yapgr to Cu.Ft.per Poun<^ , 

FIG. 145.-Specific Volume and Density of Liquid and Dry Saturated Vapor of Carbon 

Dioxide. 



476 



ENGINEERING THERMODYNAMICS 









































































<!> , 


^^ 


^o- 




















% 
















-< 


\ 


^o- 














I 

cj 
"? H 


. ft 
S" H 




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fe 


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II 

S -S 


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c3 o 
























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if 


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rt ' I 

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bc 

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x^^ 
















o-Ledoux 
<j> = Zeuner 
^=Regnault 
0= Beaton's Exp. 
A = Von Strombeck Exp 














X 




























X 




























x 


\ 




-0-0 
























\ 


r 


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h 




























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o 






























\4K, 






























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. 

V 






























5 



> en 
8 






HEAT AND MATTER 



477 



The volumes of dry-saturated steam determined from 
ompared with their pressures show that there exists 
elation of the form for steam, 

p(Vv- Fz,) 1 ' 0646 = constant = 497, 



the 
an 



tables when 
approximate 



(677) 



tvhen pressures are in pounds per square inch and volumes are in cubic feet. 
?his curve plotted to PV coordinates is called the saturation curve for the vapor, 
t is useful in approximate calculations of the work that would be done by steam 
xpanding so that it remains dry and saturated or the work required to compress 
rapor such as ammonia under the same conditions. But as the specific volume 
f liquid is generally negligible it may be written as one of the general class 



: = constant, (678) 



or which s = 1.0646 and constant = 497. 

?his curve supplies a means for computing the work for wet vapors (not too wet) 
is well as. dry, provided only that they at no time become superheated or change 
heir quality, by using for V some fraction of the true specific volume repre- 
enting the dryness. The very fact that a great volume of vapor may be 
ormed from an insignificant volume of liquid makes the saturation curve a 
jseful standard of comparison with actual expansion and compression lines for 
wet vapors. 




. 147. Comparison of Steam Expansion Line of an Indicator Card with the Satura- 
ion Line for both Dry Saturated Vapor and for Vapor Constantly Wet at the Initial Value. 



478 ENGINEERING THERMODYNAMICS 

ing for the rods, the displacement volumes of each end of the cylinder will be 
5990 cu.ins., and since the clearance volume is 4 per cent, the steam volume 
will be 239.6 cu.ins. From the left-hand card it will be seen that the cut-off 
was at point C, 16.5 per cent of stroke, hence the volume at C is (.165X 
5990) +239.6 = 1228 cu.ins. It will also be seen from the card, that the pres- 
sure at C was 73.5 Ibs. per square inch absolute. From the curves or the 
tables at the end of the Chapter 1 cu.ft. of dry steam at this pressure weighs 

1228 

.1688 Ib. and hence the weight of steam in this end of cylinder was 7^7, X. 1668 

1720 

or .1185 Ib. at cut-off. From the card it will also be seen that at the end of 
the exhaust stroke, denoted by the point D, the pressure was 30 Ibs., at which ;, 
the weight of 1 cu.ft. of dry steam is .0728 Ib., hence the weight of steam in the 



a 

cylinder was =|^X. 0728 = . 01010 Ib., and the amount admitted was .1185- 
1728 

.0101 = . 1084 Ib. 

In as much as the two ends of cylinder are identical and as the cards from 
both ends are practically the same, it may be assumed that the same weight ! 
of steam was in each end, or that .1084X2 = .2168 Ib. are accounted for by the 
card per revolution, or .2168X150X60 = 1950 Ibs. per hour. There is then 
the difference to otherwise account for, of 2600 1950 = 650 Ibs. per hour, 
which can only have been lost by condensation. 2600 Ibs. per hour is 2600 
-^(150X60X2) = . 1442 Ib. per stroke, which with the .0101 Ib. left from pre-| 
vious stroke would make .1543 Ib. in the cylinder at cut-off, and if it 
were all steam its volume would be 1581 cu.ins., denoted by point E l - 
on diagram. The ratio of AC to AE gives the amount of actual 
steam present in the cylinder at cut-off, to the amount of steam and 
water. The saturation curves CF and EG are drawn through C anc| 
E from tabular values and represent in the case of CF the volume^ 
which would have been present in the cylinder at any point of stroke had the 
steam and water originally present expanded in such a way as to keep tht 
ratio or dryness constant, and in case of EG, volumes at any point of the strok< 
if all the steam and water originally present had been in form of stean 
and had remained so throughout the stroke. Just as the ratio of AC to Al 
shows per cent of steam present at cut-off, so does the ratio of distances of an? 
points Y and Z, from the volume axis denote the per cent of steam present a 
that particular point of the stroke. By taking a series of points along th 
expansion curve it is possible to tell whether evaporation or condensation i 
occurring during expansion, In this case the ratio, 



= = .795, and IT = .86. 
AE XZ 

Hence, it is evident that evaporation is occurring since the percentage of steai 
is greater in the second case. 



HEAT AND MATTER 479 

For some classes of problems it is desirable that the external mechanical 
work be separated from the latent heat, and for this reason latent heat is 
^iven in three ways : 

(a) External latent heat, 

(6) Internal latent heat, 

(c) Latent heat total. 
The external latent heat in foot-pounds is the product of pressure and volume 

Change, or expressing pressures in pounds per square inch, 

1 

144 
External latent heat = p(7y-Fz,) (679) 

i 

this is sometimes reduced by neglecting V L as insignificantly small as it really 
s for most problems which are limited to temperatures below 400 for saturated 
apor, in which case, 

144 

External latent heat = -j-pV v (680) 

J 

In all cases 



Internal latent heat = L (Ext. Lat. Ht.) (a) 

144 

= L- 

= L-~~pV r (c) 



(681) 



Fusion and freezing are quite similar to vaporization and condensation 
i that they are constant temperature processes with proportionality between the 
mount of substance changing state and the amount of heat exchanged. They 
re different in as much as little or no volume change occurs. As there is so little 
sternal work done it may be expected that there is little change in their latent 
eats with temperature and pressure, but as a matter of fact it makes very 
ttle difference in most engineering work just how this may be, because prac- 
cally all freezing and melting takes place under atmospheric pressure. There 
oes not appear to be any relation established between heats of fusion like 
lose for vaporization that permit of estimates of value from other constants, 

direct experimental data must be available and some such are given for a 
substances at the end of this Chapter in Table XL VI. As a matter of fact 
ich laws would be of little use, and this is probably reason enough for their 
Dn-discovery. 

Example 1. Pigs of iron having a total weight of 5 tons and a temperature of 
X)0 F. are cooled by immersing them in open water at a temperature of 60 F. If 
le-half of the water is evaporated by boiling, how much must there have been originally? 
i The iron must have been cooled to the final temperature of the water, which 
just have been 212 F. Also the heat given up by the iron will be the 



480 ENGINEERING THERMODYNAMICS 

product of its weight, specific heat and temperature difference, or, considering the 
mean specific heat to be .15, 

10,000 X (2000 -212) X. 15 =2,682,000 B.T.U. 

The heat absorbed by the water in being heated, considering its specific heat as unity 
will be its weight times its temperature change and, since one-half evaporates, the 
heat absorbed in evaporating it will be half its weight times the latent heat, or 

TF[(212 -60) + X970] = 637^ B.T.U. 
These expressions for heat must be equal, hence 



Example 2. A tank of pure water holding 1000 gallons is to be frozen by meai 
of evaporating ammonia. The water is originally at a temperature of 60 F. and 
ice is finally at a temperature of 20 F. The ammonia evaporates at a pressure^ 
of one atmosphere and the vapor leaves the coils in a saturated condition. Howf 
many pounds of ammonia liquid will be needed, how many cubic feet of dry saturated! 
vapor will be formed, and how much work will be done in forming the vapor? 

The heat to be removed is the sum of that to cool the water, the latent healij 
of fusion of ice, and that to cool the ice, or for this case 

[(60 -32) +144+.5(32 -20)1x8333, 

8333 being the weight of 1000 gallons of water. Hence the B.T.U. abstract 
amount to 1,466,608. 

Each pound of ammonia in evaporating at atmospheric pressure absorbs 594 B.T.U.'. | 
as latent heat and, therefore, 2470 Ibs. are needed. At this pressure each pound 
vapor occupies 17.5 cu.ft., hence there will be 43,200 cu.ft. of vapor. At this saml*s 
pressure the volume of a pound of liquid is .024 cu.ft., so that the work done per pounj 
in evaporating the ammonia is 37,000 ft.-lbs. and the total work is 915 XlO 6 ft.-ll 

Prob. 1. How much ice would be melted at 32 F. with the heat necessary 
boil away 5 Ibs. of water at atmospheric pressure, the water being initially at 
temperature corresponding to the boiling-point at this pressure? 

Prob. 2. What is the work done during the vaporization of 1 Ib. of liquid anhydroi 
ammonia at the pressure of the atmosphere? 

Prob. 3. From the tables of properties of anhydrous ammonia check the value I 
the constant in Trouton's law given as 23.6 by Young. 

Prob. 4. As steam travels through a pipe some of it is condensed on account 
the radiation of heat from the pipe. If 5 per cent of the steam condenses how mua 
heat per hour will be given off by the pipe when 30,000 Ibs. of steam per hour at;| 
pressure of 150 Ibs. per square inch absolute is passing through it? 

Prob. 6. Brine having the specific heat of .8 is cooled by the evaporation of ammoi 
In coils. If the brine is lowered 5 F. by ammonia evaporating at a pressure of 1 1 



HEAT AND MATTER 481 

bs. per square inch gage, the vapor escaping at brine temperature, how many pounds 
f brine could be cooled per pound of ammonia? 

Prob. 6. Steam from an engine is condensed and the water cooled down to a 
emperature of 80 F. in a condenser in which the vacuum is 28 ins. of Hg. How many 
lounds of cooling water will be required per pound of steam if the steam be initially 

per cent wet? 

Prob. 7. A pound of water at a temperature of 60 F. is made into steam at 100 
bs. per square inch gage pressure. How much heat will be required for this, and what 

be the volumes at (a) original condition; (b) just before any steam is made; (c) after 
Jl the water has been changed to steam? 

Prob. 8. A sand mold weighs 1000 Ibs. and 100 Ibs. of melted cast iron are poured 
nto it. Neglecting any radiation losses and assuming the iron to be practically at 
:s freezing temperature how much of the iron will solidify before the mold becomes 

1 the same temperature as the iron? 

Prob. 9. How many pounds of ice could be melted by heat given up by freezing 
.0 Ibs. of lead? 

11. Gas and Vapor Mixtures. Partial and Total Gas and Vapor Pressures, 
/blume, Weight, and Gas Constant Relations. Saturated Mixtures. Humidity. 

)ne of the characteristic properties of gases distinguishing them from liquids, 
ind which also extends to vapors with certain limitations is that of infinite 
xpansion, according to which no matter how the containing envelope or volume 
)f the expansive fluid may vary, the space will be filled with it at some pres- 
sure and the weight remain unchanged except when a vapor is brought to 
;ondensation conditions, or the pressure lowered on the surface of a liquid 
vhich will, of course, make more vapor. A given weight of gas or vapor (within 
imits) will fill any volume at some pressure peculiar to itself, and two gases, 
wo vapors, or a vapor and gas, existing together in a given volume, will fill it 
it some new pressure which is the sum of the pressures each would exert sepa- 
ately at the same temperature (if non-miscible) . This fact, sometimes des- 
gnated as Dalton's Law, permits of the derivation of equations for the rela- 
ion of any one pressure, partial or total, to any other total or partial, in 
:erms of the weights of gas or vapor present, and the gas constants R. It 
ilso leads to equations for the various constituent and total weights in 
ierms of partial and total pressures and gas constants. Such equations sup- 
)ly a basis for the solution of problems in humidification and drying of air, in 
tarburetion of air for gasolene and alcohol engines, or of water gas for illumina- 
ion, and are likewise useful as check relations in certain cases of gas mixtures 
uch as the atmospheric mixture of nitrogen and oxygen, producer gas or gase- 
*us combustibles in general. 

Let wi, w 2 and w x be the respective weights of the constituents of a mixture; 

" w m = Sw be the weight of the mixture; 

" PI, P2, Px be the respective partial pressures of the constituents; 

" P m = SP be the pressure of the mixture; 

" Ri, R 2 , Rx be the respective gas constants; 

" R m be the gas constant for the mixture. 



482 ENGINEERING THERMODYNAMICS 

Then if wi Ibs. of one, and w 2 Ibs. of another gas or vapor at temperature 
T m occupy the volume V m cubic feet together, 

Vnfi^wiRiT*, ....... (a)} 

and (682) 

V m P2 = w 2 R 2 T m) ........ (b)\ 

whence 

V m (Pi+P2) = (wiRi+W2R<i)T m , ...... (683) 

or, in general, 

2P=2(wR)T m ......... (684) 

Or putting 

2P = P m , . .-. ......... (685) 

and 

or R m = ^^- 9 . . . (680) 



then 

P m V m = w m R m T m , . . . . . .... (687) 

so that the mixture will behave thermally quite the same as any one gas with! 
si^ch exceptions as may be due to a different gas constant R m . 
Dividing Eq. (682a) by Eq. (683) or (684) gives 

Pi = .wiRi _ wiRi I; 

P m wiR l +w 2 R 2 2(wR)' 

which gives the ratio of any partial pressure to that for the mixture in terms 
of the individual weights and gas constants. Hence 



W m R m > 



which gives the ratio of any partial pressure to that for the mixture in terms Oj 
its own weight and gas constant and those for the mixture. 

It is possible to express the ratio of weights as a function of gas constants 
alone which will permit of a third expression for the partial pressures in terms 
of gas constants without involving any weights. For two gases 



Whence 

Wi W 2 



HEAT AND MATTER 483 

But from Eq. (686) 



that 



R 2 
nd 



is the ratio of partial to total weights in terms of gas constants. On sub- 
titution in Eq. (689), 



(69) 



which graves /ie rafa'o o/ partial pressures of two gases or vapors to that for the 
I mixture in terms of the individual gas constants and that for the mixture, and 
I a similar expression can be found for more than two gases. The ratio of any 
i one partial, to the total weight can also be found from Eq. (689) in terms of its 
i gas constant and partial pressure, and the mixture gas constant and pressure, 
! from Eq. (691) in terms of the gas constants for the constituents and for 
'the mixture. This ratio of partial to total mixture weight gives the fractional 
composition by weight. 

It is sometimes necessary to know the volume relations in a mixture of two 
gases existing at the same pressure or two vapors or a vapor and gas, such, for 
example, as air and water vapor. In this case two different volumes existing 
together at a common temperature and pressure together make up a mixture 
volume equal to their sum. Using similar symbols 






(693) 



where Vi and V 2 are the volumes occupied by the two constituents respectively 
when at a mixture pressure P m and temperature T m , whence for the mixture 



T m ..... (694) 

or 

(695) 



484 ENGINEERING THERMODYNAMICS 

These Eqs. (694) and (695) are identical in form with (683) and (684) except that 
V replaces P, and V, P, so that all equations just derived also apply to volum< 
as the volume proportion will be identical with pressure proportions. F< 
convenience of reference these may be set down. 
From Eq. (688), 

g = ^, (696) 



which gives the ratio of any partial volume, to that for the mixture in terms oj 
the individual weights and gas constants. 

From Eq. (689) 

^=^^, (697) 

r fM 



which gives the ratio of any partial volume to that of the mixture in terms of its 
own weight and gas constant and those for the mixture. 



From Eq. (692) 

Vl = Rl(R2~Rm) 

V m R m (R2-RiY 



(698) 



which gives the ratio of any partial volume to that of the mixture in terms of the 
individual gas constants and that for the mixture. 

The volumetric composition of air is given by Eq. (697) or its equal 
numerically Eq. (692), and since the partial pressure of oxygen and nitrogen 
in air are 78.69 per cent and 21.31 per cent, these are its volumetric per cents. 

When one of the constituents is a vapor, all the preceding applies, provided 
the condition of the vapor is such that at the temperatures assumed it is not 
near the condition of condensation, but then the relations become more definite 
since the partial pressure of the vapor is fixed by the temperature. In practical 
work with gas and vapor mixtures the failure^of the perfect gas laws near the 
condensation condition is ignored and they are assumed to be true for the very 
good reason that there is no other way as good, to get numerical results. 

All liquids, and many, if not all solids will, if placed in a vacuum chamber, 
evaporate until the pressure has reached a certain value depending on the tem- 
perature, at which time the liquid and its vapor are in equilibrium, and evapo- 
ration may be said either to cease or proceed at a rate exactly equal to the rate at 
which vapor condenses, or more precisely, at equilibrium the weight of vapor 
in the vapor form remains constant. The weight of vapor that will rise over a liquid 
in a given space depends on the temperature and pressure which are related 



HEAT AND MATTER 



485 



Temperature,Deg. Cent. 
80 *P , 5? , 6 70 80 90 100 110 . l2jfl 

I ' H r-l r 1 "l t^-t '-I ' 1 1 I I ' r- 1 r 1 ! '. ' . 



800 



750 




50 



30 40 60 80 100 120 140 160 180 200 220 240 

Temperature,Deg. Fahr. 

148 Vapor Pressure of Hydrocarbons and Light Petroleum Distillates of the Gasolene 

Class. 



486 



ENGINEERING THERMODYNAMICS 



in the so-called vapor tension or vapor pressure tables and curves, such as, 
shown in Figs. 148, 149 and 150, for some liquid fuels or as given in the pre<; 
vious section for water. At any fixed temperature the vapor will con-i 
tinue to rise until it exerts jts own vapor pressure for the temperature 
the process being often described as evaporation without ebullition. If th( 
liquid or solid be introduced into a chamber containing dry gas the evapora-j 
tion will proceed precisely the same as in the vacuum until the pressure ha$! 



vo 



Temperature Deg. Centigrade 
30 40 50 60 70 



00 



i.O 



100 




ou 



80 



100 120 HO 

Temperature Deg. Fahrenheit 



100 



180 



200 



FIG. 149. Vapor Pressure of Heavy Petroleum Distillates of the Kerosene Class. 






risen by an amount corresponding to the vapor pressure for the temperature 
because each substance exerts the pressure it would if alone occupying th 
volume; when they both occupy the same volume the pressure will be thei 
sum and equal to the pressure of the gas alone, added to the vapor pressure fo 
the same temperature. There is one important practical condition, and tha 
is, time enough for the completion of the process of evaporation which proceed 
very slowly toward the end. F time .enough is allowed the vapor pressure 
establish itself and the gas is said to be saturated, and this is an importan 



HEAT AND MATTER 



487 



special case of gas-vapor mixtures. It is the condition in which the gas may 

be said to carry the maximum weight of vapor possible for the total pressure 





Temperature, Deg. Centigrade 
30 40 50 

J 1 1 1 I L 




70 ' 80 ' 90 ' 100 ' 10 ' 120 130 140 150 160 170 
Temperature Degrees, Fahrenheit 



FIG. 150. Vapor Pressure of the Alcohols. 

and temperature. The gas in contact with the liquid may carry less vapor 
if it has not been in contact long enough aVthe given temperature, and a gas 



488 ENGINEERING THERMODYNAMICS 

no longer in contact with the liquid may carry less, because, (a) of insufficient 
time of original contact; (6) of condensation of some it originally carried; (c) 
of a rise of temperature after leaving the liquid. To all such general cases the 
equations above apply without change, but for the special case of saturated 
mixtures they have a simpler form. 

Let P v be the vapor pressure of the liquid for temperature T, which is the 

partial pressure of the vapor in a saturated vapor gas mixture ; 
P g be the partial pressure of the gas at same temperature. 
Then for a gas saturated with vapor at temperature T, Eq. (689), 

/Weight vaporX _ /Vapor pressureX iR for mixtureX 
\Weight mixt./ " \Mixture press!/ X \ R for vapor / ' ' ' 

But according to Eq. (646) and Eq. (651), 

'R for vaporX _ /Density of mixtureX _ /Mol. wt. of mixt.X ( 
ft for mixt./ ~ \ Density of vapor / " \Mol. wt. of vapor/ ' ' 

whence 

Wy _P V m v 

w m P m m m ' ..... 

Also 

C Plfni' ' ' ' ' ; ' ' ' (7( 
and 



w v 



The presence of water vapor in the atmosphere, and problems connected with 
it, constitute a specific case of vapor-gas mixture, subject to the foregoing laws. 
This subject has been given most attention by the United States Weather 
Bureau; tables have been prepared for ready computation and for certain 
problems for which only experimental data or empiric formulas afford solution. 

Air is said to be " saturated with moisture " when it contains the saturated 
vapor of water. It might be better to say that the space is saturated since the 
presence or absence of the air has no effect upon the water, vapor other than im- 
posing its temperature or imparting heat to the water vapor, and also that the 
air retards the diffusion of water particles. The weight of saturated aqueous 
vapor per cubic foot depends only on the temperature, and not on the presence 
of air. 

If the space contains only a certain fraction of the weight of aqueous vapor 
corresponding to saturation, that fraction is called the " relative humidity " 
or degree of saturation, and the corresponding percentage, the per cent of 
saturation. If air containing saturated water vapor be cooled ever so little 



HEAT AND MATTER 489 

some of the vapor will be condensed and appear in the liquid form. If air is 
cooled at constant pressure, from a given initial condition, the degree 
of saturation approaches unity, and finally reaches that value at a temperature 
called the " dew point " corresponding to the initial condition. At this tem- 
perature the condition of saturation has been reached and any further cooling 
will cause the precipitation of liquid water, as occurs in the formation of dew, 
rainclouds or fog. 

A space or body of air carrying water vapor in smaller quantity than that 
;of saturation, in reality contains superheated steam. If the vapor density 
iand the temperature of the mixture be known, the degree of superheat may be 
[ascertained from the temperature of the mixture, and the temperature corre- 
I spending to saturated water vapor having a pressure equal to the partial pres- 
jsure of the vapor in the mixture. 

Humidity of atmospheric air is ordinarily determined by an instrument 
[called the psychrometer, which consists merely of two thermometers, one with 
| a bulb exposed directly to the air and the other covered with a piece of wick 
[which is kept moist with water. The two are mounted together so that they 
[can be whirled or swung about in the air until a stable condition has been 
[reached. The dry-bulb thermometer should record the temperature of the air. 
[The wet-bulb thermometer will record something lower* than the air tempera- 
jture, dependent upon the rate at which evaporation takes place, since the process 
jof evaporation abstracts heat. Were there no other influence, this process of 
[evaporation would continue till the temperature of the wet bulb became that 
I of the dew point. The temperature of the wet-bulb thermometer never falls 
[to the dew point, however, because of conduction of heat between the cold 
bulb and the warmer surrounding air. From extensive experiments conducted 
I by the U. S. Weather Bureau, Professor Ferrel has devised the following formula 
jfor the vapor pressure, h in ins. of mercury corresponding to given readings 
{of the wet- and dry-bulb thermometers, U and t w degrees F. 



(704) 



where h b is barometric height in inches, after applying all corrections, and h' is 
pressure of saturated water vapor, in inches of mercury, corresponding to the 
temperature t w . 

The relations shown by this formula are expressed graphically in much 
more convenient form in the curves of Figs. 151 and 152, devised by 
Prof. H. L. Parr. The use of the curves is best illustrated by an example: 
f the dry-bulb reading is 75 F. and the wet-bulb 65 F., find the dew point. 
The difference of wet- and dry-bulb temperatures is 10. From 10 at the top of 
;he diagram (B) Fig. 151 project downward, and from 75 air temperature at the 
.eft of diagram project to the right to the intersection, where the dew point is 
Iread by interpolation between the contour curves at (C) to be 59.5 F. These 
?urves are drawn for a barometric pressure of 29.92 ins. (standard) and will 



490 



ENGINEERING THERMODYNAMICS 



IOC 



Difference in Temperature: Wet and Dry Bulbs: Degrees Fahrenheit 
2 4 6 8 10 12 14 1 1 




2 34 5 6 7 89 11 

Difference in Temperature: Wet and Dry Bulbs: Degrees Centigrade 

FIG. 151. Relation between Wet and Dry Bulb Psychrometer Readings and Dew Point for 

Air and Water Vapor. 



HEAT AND MATTER 



491 



\ \ \ \ l\ l\ IX IX 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



\ 



X 



& 



X 



X 



X 

X 



100 



90 



80 



70 



60 



be 

& 

:> OJ 

I 

I 

a 



40 



30 : 



10 20 30 40 50 60 70. 

Degree of Humidity, Per Cent 

FIG. 152. Relation between Humidity and Weight of Moisture per Cubic Foot of 

Saturated Air. 



492 ENGINEERING THERMODYNAMICS 

not apply correctly, when the barometer is not equal to this, though with fair 
approximation, so long as the difference in barometer is not great. Where there 
is much departure the original the formula must be used. Fig. 152 gives 
weight of aqueous vapor per cubic foot of mixture, in grains (7^5 lb.) and 
also the degree of humidity. The temperature of the dew point 59.6 F, is 
located at (C") on the right-hand side of Fig. 152. Interpolation between the 
ends of the contours for weight, gives 5.6 grains per cubic foot. On the same 
scale the temperature of the air, F., is represented at point (A) 75, project- 
ing to the intersecting point D and down to the bottom of the diagram gives 
on the scale for degree of humidity, 60 per cent. 

Example 1. By means of the relation of gas constants find the proportion of nitrog 
and oxygen in the air. 

R for nitrogen is 54.92 and for oxygen 48.25 and for air 53.35. From Eq. (698) 



V m R m (RN-RoY 
which, on substituting the above values for R N , RQ and Rm gives 

j^ 48.25 (54.92-53.35) 

V m "53.35 (54.92-48.25) ' 

or air is 21.3 per cent oxygen by volume. 

Example 2. At what temperature will air containing ^ lb. of water per pound of l 
dry air at atmospheric pressure be saturated? 

If the vapor pressure be known, the temperature may be found from tables, i 
From Eq. (701) 



Wg 

or 



For air m g = 28.88 and for water m p = 18, 
hence substituting those values 



, and P c +P,=760mm. of Hg. 



(760-P c )(28.88x.5) 
18x1 



or 



which corresponds to a temperature of 172 F. 



HEAT AND MATTER 493 

Example 3. A pound of alcohol requires 9.06 Ibs. of air for a proper combustible 
mixture for gas engines. At what temperature will these proportions contitute a 
saturated mixture? 

From Eq. (701) 

^ 
~ 

For alcohol m -46, for air m g = 28.88, and P V +P =7W mm. of Hg for atmospheric 
pressure. Substituting these values in the above equation 



(760-P,)28.88 
46X9.06 



From the curve of vapor tension of alcohol, the temperature corresponding to 49 mm. 
of Hg is about 72 F. , 

Prob. 1. Air at 80 per cent lumidity, atmospheric pressure and 70 F, is cooled 
to 40 F. How much water will be thrown down per 1000 cu.ft. of moist air. 

Prob. 2. The same air is compressed adiabatically to five atmospheres, and again 
cooled to 40 F. at this pressure. How much moisture per 1000 cu.ft. of moist air 
fwill be separated out when the tenperature becomes 70 F, and bee much at 
J40 F. 

Prob. 3. What will be the weight of water in a pound of air and water vapor if 
[the value for R for the mixture is taken as 55.25, for air as 53.35 and for water vapor 
las 91? 

Prob. 4. At what temperature will air containing its own weight of water vapor 
|be saturated at atmospheric pressure? 

Prob. 5. An internal combustion engine uses a saturated mixture of air and gasolene 
jvapor in which ratio of air to gasolene is 15.3. Considering the gasolene to be hex- 
lane, at what temperature will the mixture be? 

Prob. 6. Should kerosene regarded as decane, C 10 H 2 2, be substituted for gaso- 

e, in the above problem what would be the change in temperature of mixture, 
uming it still to be saturated? 

Prob. 7. Air containing moisture equal to one per cent of the weight of the air 
ijilone is at a temperature of 150 F. How much is the water vapor superheated? 
iVhat is the humidity? 

Prob. 8. The reading of a dry bulb of a psychrometer is 90 F. and of the wet 
)ulb 70 F. By means of curves of Figs. 151 and 152, find the dew point, relative humid- 
ty, and grains of water per cubic foot of air. 

12. Absorption of Gases by Liquids and Adsorption or Occlusion by Solids. 
Relative Volumes and Weights with Pressure and Temperature. Heats of 
Absorption and of Dilution. Properties of Aqua Ammonia. In the attainment 
)f high vacuua in steam condensers, separate removal of considerable quantities 
df non-condensible gases is found necessary by means of dry vacuum pumps, 
i fact that proves in a practical way the freedom with which the boiler water 
lad absorbed gases. These gases for otherwise pure water are carbon dioxide 
iind air, probably mainly air, but may include many others, notably the 



494 ENGINEERING THERMODYNAMICS 

products of organic decomposition especially when condensing water is taken 
from the neighborhood of sewers, as is generally the case when power plants 
are located on city water fronts. To a very much greater extent, however, 
is ammonia soluble in water, the latter being capable of taking up about 1000 
volumes of ammonia at and one atmosphere, against about 30 volumes of air, 
and one-fiftieth of a volume of hydrogen. It is the freedom of solution of am- 
monia in water that makes the process useful as a means of removing 
anhydrous ammonia from the cooling coils in mechanical refrigerating plants, 
as a substitute for the mechanical removal by piston compressors. 

In all cases the solubility of gases in liquids decreases with rise of tem- 
perature, a fact associated with the separation of gases from boiler feed- 
water during their heating in feed-water heaters, economizers and the boiler 
itself. This property is also depended upon to free the aqua ammonia that 
has absorbed its ammonia charge from the cooling coils, of the amount so 
taken up, by heating the solution in a separate chamber from which the am- 
monia vapor escapes to the ammonia condenser to become liquid anhydrous 
while the weak liquor returns to the absorber for a new charge. To permit of 
the calculation of the quantity of liquor to be circulated, in order that a given 
amount of anhydrous ammonia may be absorbed from the cooling coils and 
delivered later by heating to high temperature to a condenser, requires accu- 
rate data on the maximum possible ammonia content of solutions at various 
temperatures and pressures. Rise of temperature always will reduce the 
gas content of the solution if originally saturated, but the volume dissolved 
is independent of pressure for slightly soluble gases like nitrogen or hydro- 
gen, the weight dissolved, of course, being greater and directly proportional to 
pressure at a given temperature by reason of the increased density. 

This law of independence of volume and pressure or proportionality 
of weight to pressure, is known as Henry's law, and is hardly true for gases 
as freely soluble as ammonia, probably due to some action between water and 
the gas, equivalent to that studied by Thomsen for solids, which tend to form 
hydrates of various kinds. For such gases as are soluble by weight in propor- 
tion to pressure, it is not the total pressure of the solution that is significant, 
but the partial pressure of the gas alone, so that the amount of mixed gases 
like air dissolved in water will depend, for the oxygen part, on the specific 
solubility of oxygen and its partial pressure in the air, which is approximately 
one-fifth that of the air, and the same is true for the nitrogen. Thus, in examin- 
ing the solubility conditions for ammonia in water, while in practice the total 
pressure only is known, it is to the separate pressure of the ammonia that the 
amount dissolved must be referred in any attempt to establish a law of 
relation. 

Just as gases dissolve in liquids, so are they absorbed by solids, though in this 
case the process is described as one of adsorption or occlusion. This phenomenon 
is now being studied in connection with coal, which it is found more or less 
freely absorbs air, the oxygen of which under comparatively low temperatures 
unites with the coal causing spontaneous combustion if the heat is conserved 



HEAT AND MATTER 



as in 
Most 



a pile or its liberation in any way accelerated by heating or otherwise, 
investigations of adsorption or occlusion have been made with charcoals, 




32 50 



70 



110 130 150 170 190 

Temperature in Degress Fahrenheit 



210 230 



TIG. 153. Ammonia-water Solutions, Relation between Total Pressure and Temperature 

(Dotted Lines Mollier Data). 

,he more dense varieties of which have greater adsorptive power than others. 
Hie quantity of different gases adsorbed is believed by Dewar to be the same 
n volume per unit of charcoal when each is held at its own condensation tern- 



496 



ENGINEERING THERMODYNAMICS 



perature. The quantity increases with rise of pressure but not in proportion, 
and decreases rapidly with rise of temperature and a curve showing the tem- 



130 



120 




10 ll. 15 20 23.3925 30 

Per Cent by Weight of Ammonia in Solution. 



FIG. 154. Ammonia-water Solutions, Relation between Total Pressure and Per Cent NH 

in Solution. 

perature and pressure at which equal volumes are adsorbed is similar in form 
to a vapor tension curve. 

In the establishment of the properties of aqueous solutions of ammonii 



HEAT AND MATTER 



497 



progress was for many years very slow and the early experimental data of 
Abegg and Rosenfeld, 1903, for weak solutions and low pressures, that of Roscoe 
and Dettman, 1859, for highly concentrated solutions under low pressures at 



350 



325 



300 



275 



250 



S 225 

| 

200 



150 

I 
| US 

100 
75 
50 
25 



^X 




Bill 

uare-In 









5 10 11.8 15 20 23.3925 30 -'& & 

Percent by Weight of Ammonia in Solution 

FIG. 155. Ammonia-water Solutions, Relation between Temperature and Per Cent NH 8 

in Solution. 

32 F., together with that of Sims, 1862, for four temperatures from 32 F. 
to 212 F., failing to check and having many gaps, has been more recently supple- 
mented by Perman, 1903, and Mollier, 1908. With these new data it has been 
Possible to graphicajly fill in the data of the unexplored region with reasonable, 



498 ENGINEERING THERMODYNAMICS 

though not yet satisfactory accuracy. The results are given in three diagrams, 
grouping the three variables of pressure, temperature and concentration in 
pairs, Fig. 153, giving pressure-temperature, Fig. 154, pressure-concentration 
and Fig. 155, temperature-concentration, from which can be read off with reason- 
able accuracy any quantity needed in calculations and from which Table 
LI, has been prepared. In this table the lower numbers are new, and the 
upper those as given by Starr several years ago and since used by engineers 
engaged in refrigeration, as standards. 

These data refer to he equilibrium conditions of the solution, and in using 
them for practical problems care must be taken to avoid applying them to other 
conditions, for example to solutions that are not homogeneous, or in which 
there has not been sufficient time for the establishment of equilibrium. Over 
the surface of such solutions there exists a mixture of water vapor and 
ammonia, each exerting its own partial pressure and the sum of the partial 
pressures making up the pressure of the solution; it must not be assumed, 
however, that the partial pressures are those of the pure substances, as this is 
a true solution and not a simple mixture. Moreover, there is no certainty that 
it is always a solution of just ammonia in water, as there is a reasonably good 
possibility that hydrates of ammonia may form, which would further complicate 
the relative pressures of the two constituents. 

It is from Perman that the most accurate data on the composition of the 
equilibrium vapor mixture have come, and he calculated the partial pressures 
from the composition of the mixture determined by analysis and using Dalton's 
law, 

Partial pressure of ammonia _ Volume of ammonia 
Total pressure "Total vapor volume' 

No success has yet been met with, in attempting to express a general relation 
between partial pressures and the two variables, pressure and temperatures, 
so Perman's values are given as found in Table LII, and for the unex- 
plored region it is not possible to do better than make a guess at a needed value. 
As a check on the Perman values and to assist in estimating needed values for 
other ranges the sum of the two partial pressures is given and compared with 
the accepted values for the total pressure, and it should be noted that as the 
partial pressure relations give the volumetric composition of the vapor mix- 
ture, it is unfortunate that the data are not extended to at least 35 per cent 
concentration to cover the solutions in the generator of the absorption refriger- 
ating machine, from which both ammonia and water-vapor are discharged 
in as yet unknown relative amounts and which must be separated as completely 
as possible before condensation. 

Any change in the ammonia content of a solution is thermal in character 
and is, therefore, accompanied by heat changes. When water absorbs ammonia 
heat is liberated, as is also the case when ammonia in solution is diluted with 
more water, the latter being really a further absorption in the fresh water of 
the ammonia already contained in its solution being diluted by it. Likewise, 



HEAT AND MATTER 499 

when ammonia is absorbed by an ammonia solution heat is also liberated, 
but heat is absorbed by a solution from which ammonia is escaping, as in 
evaporation of liquids. There are three cases of the exothermic process, each 
with an endothermic inverse and these are: 

(a) Absorption of ammonia by pure water. 

(6) Absorption of ammonia by ammonia solution. 

(c) Dilution by water of an ammonia solution. 

Data on the amounts of heat liberated in these cases are not sufficient 
to establish firmly any general law of change, but are sufficient to give an 
approximation. The first important fact in this connection is that the heat 
liberated per pound of ammonia when pure water absorbs ammonia depends on 
the amount of water. One pound of ammonia absorbed in a little water gives 
out a little heat, more is liberated when more water is present, but when the 
amount of water is large, put at 200 times the weight of ammonia by Thomsen, 
the heat of absorption is constant and does not increase beyond this point. 
It may easily be, however, that the point is reached with fifty water .weights, or 
that some heat continues to be generated for any amount of water to infinity, 
but so small in quantity as to be impossible to measure in the great weight 
of liquid present. For example, if one B.T.U. were liberated in 100 Ibs. 
of solution, the temperature rise would be somewhere near to Vioo F., and to 
detect this in the presence of radiation and conduction influences and make 
allowance for the heat of stirring would be difficult. 

For engineering purposes, however, it may be accepted that the heat of absorp- 
tion of a pound of ammonia is constant if the weight of water is large, and its value 
was fixed at 926 B.T.U. per Ib. by Favre and Silberman and later redetermined 
by Thomsen at 8430 calories per gram molecule of NHa, which is equivalent to 

9 8430\ 

X ) = 893 B.T.U. per pound. This value is accepted and defined as the 

5 17 / 

heat of complete absorption for the want of a better term, and in view of the 
desirability of distinguishing it from the heat of absorption in small amounts 
of water or in solutions already containing appreciable amounts of ammonia, 
which will be designated as heat of partial absorption. 

Experiments have further established another important relation between 
the heats of dilution of solutions and their original strength. According to 
this, solutions behave like ammonia itself with respect to pure water and lib- 
erate a little heat when a little water is added, more with larger amounts, 
attaining a constancy for very large amounts of water. Thus a solution of a 
given ammonia strength, it may be assumed, will always liberate the same 
amount of heat when diluted with water, if the total amount of water after 
dilution exceeds two hundred times the weight of ammonia present or there- 
abouts. This heat per pound of NHs contained will be designated as the 
heat of complete dilution, and defined as the heat liberated when a solution 
containing 1 Ib. of ammonia in a given amount of water is completely 
diluted with water, or brought to the condition of 200 Ibs. or more of water per 
pound of ammonia contained. 



500 



ENGINEERING THERMODYNAMICS 



There is a rational relation between these three heats, that of complete 
absorption, which is a constant, that of complete dilution, which depends only 
on the original ratio of ammonia to water, and that of partial absorption, 
which is a function of the character of solution receiving it, or if pure water 
the amount. This relation is 



J Heat of 
complete 
I absorption . 



(Heat of partial absorption) =gg3 
[ (+Heat of complete dilution) J 



]b 



Numerical values for heats of partial absorption are entirely lacking, but 
Berthelot has given ten values for the heats of complete dilution for solutions 
containing from 1 Ib. of water per pound of ammonia, to a little over one 
hundred, but at only one temperature, 57 F. Up to Thomson's time these 
figures seem to have been the sole dependence for engineering calculation; 
he, however, added three more figures for more concentrated solutions, giving 
not the heats of complete dilution, but those of partial dilution, that is, the 
heat liberated when the water content is increased from one original value to 
three different final values not corresponding to the state of complete dilu- 
tion. However, these may be used to check and correct the Berthelot figures 
and are especially useful because they cover the doubtful range of his deter- 
mination and the range of ammonia strengths. 

Berthelot's results are given in the following Table XXV and plotted in 
Fig. 156, from which he derives a general law of relation given by the follow- 
ing formula Eq. (706), which is also plotted to show its agreement with the 
points, and which is an equilateral hyperbola asymptotic to axes of H and w. 



TABLE XXV 

BERTHELOT'S DATA ON HEATS OF COMPLETE DILUTION OF AMMONIA 

SOLUTIONS 



Original Solu- 
tion, 1 Lb. Am- 
monia in (w) 
Lbs. Water. 


When Completely Diluted gives 


Original Solu- 
tion, 1 Lb. Am- 
monia in (to) 
Lbs. Water. 


When Completely Diluted gives 


1.04 


136] 




3.76 


34] 




1.06 


134 




6.11 


22 




1.13 

1.98 


124 
51 


B.T.U. per Ib. ammonia 


10.06 
57.39 


2 



B.T.U. per Ib. ammonia 


3.18 


41 J 




116.47 


OJ 





142.5, 



Heat of complete dilution, B.T.U. per Ib. NH 3 = (Berthelot) . . (706) 



w 



The agreement, it will be noted, is not very good for larger values of w than 
4 or 5, which is unfortunate, as commercial ammonia lies between one part 
ammonia to nine parts water, or w = 9, and one part ammonia to 39 



HEAT AND MATTER 



501 



parts water, or w = 39. Nevertheless engineers have been using this formula 
in these doubtful ranges for some years. 

By means of the few but probably accurate figures given by Thomsen and 
experimentally determined by him it is possible to check this practice. He 
measured not the heat of complete dilution as did Berthelot, but the heats of 
partial dilution, and the manner in which his data merge into those of Berthelot 
make the combined results of doubly great value because of the difference in 
method. Thomsen added a limited amount of water to a solution of ammonia 
containing 3.39 Ibs. water per pound of ammonia and measured the heat, 
which, of course, was not the heat of complete dilution, with the following 
results: 

THOMSEN'S DATA ON HEATS OF PARTIAL DILUTION OF AMMONIA 

SOLUTIONS 



Original Solution. 
,' 1 Lb. Ammonia in (w) Lbs. Water. 


Final Solution. 
1 Lb. Ammonia in (w) Lbs. Water. 


B.T.U. per Lb. Ammonia Absorbed 
by the Ammonia Solution. 


(3.39 
10= ] 3.39 
13.39 


T19.27 

w = \ 29.86 
[56.33 


+34 
+37 
+40 



These results have been fitted into those of complete dilution by the relations 
of Eqs. (707) and (708), and by the introduction of one assumption. 



Heat of complete dilu- 
tion of original am- 
monia solution per 
Ib. NH 3 



Heat of partial dilution 
from original to some 
greater water content 
per Ib. NH 3 . 



Heat of complete 
dilution of the 
new solution per 
Ib. NH 3 . 



(707) 



or 



j [Heat of complete dilu-1 ("Heat of complete dilu-1 
{ tion of new solution [ = \ tionof original solution \ - 
[ per Ib. NH 3 J I per Ib. NH 3 . 



Heat of partial dilution! 
from original to final f (708) 
solution per Ib. NH 3 . J 



If the heat of complete dilution of the original solution containing 3.39 Ibs. 
water per pound ammonia- be taken from the Berthelot equation its value is 
42, so assuming this to be correct it may be introduced in Eq. (708) as a constant, 
.giving with the Thomsen figures the following: 



Heat of complete dilution | 
of new solution, per Ib. [ 
NH 3 . 



{Heat of partial dilution j 
from original to final r 
solution per Ib. NH 8 . J 



Heat of complete dilution per pound NH 3 with 

(10 = 19.27) =42-34 (Thomsen) = 8 ; 
(w = 29.86) =42-37 (Thomsen) = 5 ; 
(to = 56.33) = 42 - 40 (Thomsen) = 2. 



502 



ENGINEERING THERMODYNAMICS 



These three new points are also plotted and agree well with the Berthelot 
equation, better even than the original Berthelot points themselves, so that 
Thomsen's partial dilution figures seem to confirm Berthelot's complete dilution 
data and the curve of Fig. 156 and Eq. (706) may be taken as truly represen- 



150 




5 10 15 20 25 30 

Pounds of Water Per Pound of Ammonia 

1 K;. 156. Heat of Complete Dilution of Ammonia Water Solutions, by Excess Water. 

tative of the heat of complete dilution of ammonia solutions and indirectly 
of course, heats of partial dilutions as well. 

Heats of absorption are more often needed in practical problems than heat 
of complete or partial dilution, but these heats follow on the assumption that th' 
heat of complete absorption, must be equal to the sum of the heat of partial absorp 
tion, and the heat of complete dilution of the solution so formed. 



HEAT AND MATTER 



503 



Hence 



Heat of partial absorption 
in w Ibs. water B.T.U. 
per Ib. NH 3 . 



Heat of complete 
absorption in ex- 
cess water B.T.U. 
per Ib. NH 3 . 



[ Heat of complete dilution by 
J excess water of solution 

with w Ibs. water per Ib. 

NH 3 , B.T.U. per Ib. NH 3 . 



or 



( Heat of partial absorption in water ^ 
I B.T.U. per Ib. NH 3 absorbed / " 



142.5 

w 



(709) 



800 



a 

I 



400 



200 



Heat of par 



ial absorption 

112.5 
W 



Heat of complete 

U8.5 

V 



Hea of complete abs< 



dilutioi 



rption 



10 



15 20 25 30 

W =Lbs. Water per Ib. NH 3 



FIG. 157. Ammonia-water Solutions, Relation between Heats 

f Partial Absorption j 

of \ Complete Absorption \ Shown Graphically. 
I Complete Dilution / 

It is interesting to note that the relation between these three quantities, heat 
of partial absorption in limited amounts of water, heat of complete absorption 
in excess water and heat of complete dilution of the solution in excess water, 
can be shown graphically in Fig. 157, where the designations are self- 
explanatory. 

Most important are the heats liberated when ammonia is absorbed not 
by water but by weak ammonia solutions themselves, and these heats of partial 
solution of ammonia in ammonia solutions are obtainable from the data already 
established by a comparatively simple relation. In this case there are two 
different solutions in question, an original one which becomes the second one 
on receiving more ammonia. If the water received all the ammonia contained 
in the second solution a certain quantity of heat would be liberated and it must 
be equal to the total amount liberated by absorption of the first ammonia in 
the water, and by the absorption of the second ammonia in the resulting solu- 



504 



ENGINEERING THERMODYNAMICS 



tion, whence this last quantity is obtainable by differences between the heats 
of partial solution of ammonia in water alone. 



Let w =lbs. water per Ib. ammonia in original solution which, therefore, consists of 

w + 1 Ibs. solution, 1 Ib. of ammonia and w Ibs. of water. 
A =lbs. ammonia added per Ib. ammonia already present, making new solution 

containing w Ibs. of water and A+l Ibs. of ammonia or Ibs. water 



Then 



per Ib. ammonia in w+A +1 Ibs. of solution. 



A+l 



(Heat of partial absorption of 1 1^9 t- 

original 1 Ib. ammonia in w [ =893- 
Ibs. water, B.T.U. 



fHeat of partial absorption of 1 

\ all (A +1) Ibs. of ammonia I 

in w Ibs. of water, B.T.U. J 



w 






Whence 

Hjeat of partial 
absorption of 
A Ibs. NH 3 by 
solution con- 
taining 1 Ib. 
NH 3 in w Ibs. 
water,B,T.U, 




(710) 



Therefore 



Heat of partial absorption of A Ibs. 
NH 3 by solution containing 1 Ib. 
NH 3 in w Ibs. water B .T.U.per Ib. 
NH 3 



=893- 



(A +2) 



(711) 



As ammonia solution strengths are often given in terms of per cent of 
ammonia present by weight and the heat of absorption in terms of changes in 
the per cent strength, the following conversion factors are useful : 

100 
Per cent ammonia in original solution = Si = , 

IP-HI 



Per cent ammonia in final solution =82 = , 



Whence 



100- 
Si 



-, (a) 



ioo/ 2 -i\ 
'' Si Vioo-ftx- 



(712) 



HEAT AND MATTER 505 

These on substitution in Eq. (711) give the heat of absorption per pound 
of ammonia absorbed to change the per cent NHs from Si to $2. 

i Heat of partial absorption of 1 Ib. ) / <? \ 

of NH 3 in a solution containing V = 893- 142.5 (^^r-f- ~-^-). . (713) 
( lib. NH 3 in w Ibs. water, B.T.U. ) \ 1C ~ bl 

When, however, the absorbed ammonia is to be driven off from the solution by 
heating it, the discharge will consist partly of ammonia vapor and partly water 
vapor, so the heat of liberation of a given amount of ammonia from solution will 
be in practice that for the ammonia and equal to what would be liberated by its 
absorption, but also in addition the heat of vaporization of the water vapor. 
When, as in absorption refrigerating machine generators, the discharged 
vapors meet incoming solution and are thereby partly condensed, prac- 
tically all except perhaps 2 per cent of the heat of vaporization of water vapor is 
returned and the net heat of ammonia liberation is not materially different 
from the value for absorption. If this is not done a correction must be intro- 
duced for the water vapor. 

To assist in the solution of problems on the amount of non-condensible 
ajases to be handled by dry vacuum pumps serving steam condensers, and on 
she change in the composition of gases when scrubbed by water in the course 
)f cooling and cleaning after manufacture, a table of solubilities of various gases 
n water is added at the end of this Chapter in Tables LIII and LIV, as compiled 
'pom various sources and reported in the Smithsonian physical tables. The 
lumbers in the tables are volumes of standard gas, that is, gas measured at 
2 F. and 1 atm. pressure, per unit volume of water, though they are at a 
Afferent volume as absorbed or when absorbed at the temperature given, 
jrhe pressure of the solution is in every case 29.92 ins. Hg., absolute and this 
the combined pressure of both the gas and the water vapor. 

Example. In the absorber of an ice machine of the absorption type, a weak solu- 
on of ammonia in water takes up the ammonia vapor coming from the refrigerating 
ils, the heat found being removed by water. The weak liquor, as it is called, is con- 
nuously supplied and the rich liquor continually pumped away to the generator 
here, by heating, some of the ammonia vapor is driven off to the condenser. As- 
iming the action in the absorber to be merely one of ammonia dissolving in a weak 
>lution and that no water vapor leaves the generator, what will be the heat produced 
the absorber and needed in the generator per pound of ammonia for the fol- 
wing assumed conditions: 

Weak solution, 15 per cent NHs; strong solution, 30 per cent NHs; 
temperature in absorber, 80 F. 

From Eq. (713), the heat of absorption per Ib. of ammonia absorbed will be 



506 ENGINEERING THERMODYNAMICS 

where Si and 2 are the per cents of ammonia in weak and rich solutions, respec- 
tively, or 

Q =893-142.5^+-^) =807. B.T.U. per Ib. NH 3 absorbed 
\8o 70 / 

Prob. 1. Ammonia is being absorbed by water at a temperature of 100 F. The 
solution contains 10 per cent of ammonia. If the total pressure is 15 ins. of Hg, what 
part of this is due to foreign gases, what part to ammonia and what part to water 
vapor? 

I^Prob. 2. How many cubic feet and how many pounds of the following gases can 
be separately dissolved in 1000 gallons of pure water at atmospheric pressure and a 
temperature of 50 F.? Air, carbon dioxide, and hydrogen. How would the results 
be changed if the pressure were doubled? If the temperature were doubled? 

Prob. 3. When either water or ammonia is added to an ammonia solution, heat 
is evolved. Explain why this is so. 

Prob. 4. Ammonia is being absorbed by a stream of running water, there being 
5 Ibs. of water per pound of ammonia. What will be the heat developed per poun 
of ammonia liquor formed? 

Prob. 5. Ten pounds of the above solution receives an additional pound of am 
monia. How much heat is generated by this action? 

Prob. 6. A solution containing 10 per cent of ammonia receives an addition o 
another 10 per cent. What was the amount of heat developed per pound of ammom 
and per pound of solution when the second portion of the ammonia was added? 

Prob. 7. The pressure in the absorber of an ammonia absorption machine is one a 
mosphere and the temperature is 80 F. What is the maximum per cent of am 
monia which can be absorbed by the water? 

Prob. 8. The generator is working under a pressure of 125 Ibs. per square inc 
gage and the heat is supplied by a steam coil in which the pressure is 30 Ibs. per squar 
inch gage. What per cent of ammonia will be left in solution after passing throug 
the generator and about how much steam must condense per ib. NHs discharged? 

13. Combustion and Related Reactions. Relative Weights and Volume 
of Substances and Elements, before and after Reaction. Not only ma 
matter assume the three states of solid, liquid and vapor separately, in pair 
simultaneously, or even all three together with various accompanying or causa 
temperature, pressure, or heat content, conditions, without chemical chang 
of the matter itself, as already discussed, but matter itself may change in kind 
As Mellor puts it " matter appears to be endowed with properties in virtu 
of which two or more dissimilar substances, when brought into close contact 
give rise to other forms of matter possessing properties quite distinct from th 
original substance." " The process of change is called a chemical reaction. 
Chemical changes are assumed to be characterized by molecular rearrangemen 
according to which molecules of elements may divide into atoms and th 
separated atoms of one combine with those of another element, to form a molecul 
of a new substance to be called a compound. Similarly, the molecules of con 
pounds may split and reassociate, part of one, with part of another, to mak 
a new compound or a single compound may split up into its elements whic 



HEAT AND MATTER 507 

remain separated, the last case being generally termed dissociation. All three 
classes of change of substance are classifiable as chemical reactions, and there is 
really no very rigid line to be drawn between the sub-classes of reaction known 
as combination and dissociation except when applied to the same substances, 
in which case one is the reverse of the other. The complete or partial destruc- 
tion of a substance as such is commonly termed decomposition as, for example, 
when hydrocarbon constituents of coal volatile, or liquid fuel, are changed by 
excessively high temperature into free carbon, and other hydrocarbons or 
even free hydrogen. 

Every chemical change whether of combination or dissociation is accompanied 
by a heat change of the system or group of substances. When the reaction 
is such that heat is set free tending to raise the temperature of the reacting 
mass unless it be carried away as liberated, the reaction is classed as exothermic. 
On the contrary, those reactions that are accompanied by heat absorption, 
tending to lower the temperature, unless heat be added to supply the absorp- 
tion, are classed as endothermic. 

It appears then that every reaction tends to change the temperature of the 
system, causing it to rise or fall according as the reaction is exothermic or 
endothermic, except in the one case where several reactions occur simulta- 
neously, in which all the exothermic, set free just enough heat to supply what 
is required for the endothermic. 

The most important reaction in engineering is combustion, defined as the 
chemical reaction between fuels and the oxygen of the air, which is exothermic 
or heat liberating, and the source of practically all the heat used in engines for 
conversion into work. Combustion is often classed as an oxidation process, and is 
thus distinguished from another important engineering class of related reactions, 
reduction or the reverse of oxidation, the most prominent case of which is 
the change from carbon dioxide to carbon monoxide in gas producers, and in 
a lesser degree in furnaces, according to which there is a reduction of oxygen 
content per molecule, and which process is endothermic. The formation of 
carbon monoxide directly from carbon by its oxidation, is sometimes defined 
as partial combustion of carbon or incomplete oxidation because the product, 
carbon monoxide, may further oxidise or burn to carbon dioxide by taking 
up more oxygen. The substances produced by the partial combustion or partial 
oxidation of one fuel element may also be considered as the result of the dis- 
sociation or reduction or deoxidation of the substances produced by its complete 
combustion. Ordinarily, partial combustion and reduction are considered 
as reverse processes producing the same substances by (a) exothermic reaction 
of the primary substance in partial combustion and (6) the endothermic deoxi- 
dizing reaction of the products of complete exothermic reactions of the same 
primary substances. It is also common to consider only the reaction of a sub- 
stance or so-called fuel element with oxygen, as combustion and other processes 
whether involving oxygen or not, as related reactions. Thus, carbon combining 
with oxygen to form carbonic acid is complete, and carbon combining with 
k oxygen to form carbon monoxide incomplete combustion of carbon, while carbon 



508 ENGINEERING THERMODYNAMICS 

monoxide and steam reacting can hardly be considered as combustion, and is 
best classified as a related reaction. 

The number of elements entering into combustion and related reactions 
is small, but the number of possible substances and reactions between them is 
amazingly large, and the prediction of just what reactions will take place between 
various groupings or mixtures of these substances extremely difficult and in some 
cases quite impossible. The study of possible reactions has become the province 
of physical chemistry, especially when the conditions controlling the result 
are also subjects of study. These conditions include the temperature, pressure, 
electrical state and the mutual relation of the elementary compounds present, and 
the relation of these various conditions to the primary and resultant substances, 
and the intermediate, successive, simultaneous or parallel reactions constitute the 
subject matter of the study of chemical equilibrium. If chemical equilibrium 
were better understood than it is, it would be possible to predict the resultant 
from primary substances for specified conditions, but at the present time this 
is impossible even though some of the greatest scientists the world has ever 
known, have devoted their lives to the study and many volumes of specific 
results have been published. Even if the exact prediction of the direction in 
which reaction would proceed in an ordinary complex system and the extent 
to which it would go in that direction, were made possible by a more complete 
thermochemistry than now exists, it would not be of much use in engineering 
because it is seldom possible to define the conditions that are present or to be met. 
For example, in gas producers, solid coal, air and steam are the primary materials 
and the product or result of their mutual reaction is a combustible gas. Engi- 
neers would like very much to be able to predict and control the exact com- 
position of this gas, but this is not possible nor will it in all probability ever be 
possible, because it would first be necessary to fix the chemical and thermal 
character of the coal, air and steam, to fix their relative quantities, to maintain 
an absolutely uniform fuel bed as to size, porosity and, quality with some 
other conditions equally elusive. 

It must be understood, therefore, that while the possibility or even probability 
of certain reactions taking place may be known, it is quite impossible to 
predict just what will happen, or what products will result, when a given 
group of primary substances mutually, react so that many important problems 
of combustion in boiler furnaces and gas producers cannot be solved except by 
approximation. 

The approximation takes the form of a calculation which is exact, based on 
an hypothesis which does not represent the facts of the case. In other words, 
engineering calculations about combustion are always to be prefaced by a 
statement that certain substances are going to change completely or within 
a given degree to certain others, whether they will or not. Furthermore, the 
substances must be defined chemically by their symbols, and then will it be 
possible to calculate the relative weights and volumes of the various substances 
that can so react, the corresponding relative weights and volumes of the products, 
and the heat liberated or absorbed, but not otherwise. It is quite important 



HEAT AND MATTER 509 

that too much confidence be not put in these results, which are no more correct 
i than is the assumption of what substances are to be formed. 

From the study of chemical equilibrium a few principles of guidance have 
ibeen developed that help a little but not very much. For example, Van't 
! Hoff's " law of movable equilibrium " says that when the temperature of a system 
| is raised that reaction takes place which is accompanied by absorption of heat 
[and conversely. Another similar law is to the effect that a rise of pressure in 
jja system in equilibrium causes that reaction that is accompanied by a reduction 
jof volume. There are more of this sort but they are entirely too general to make 
[at possible to avoid the procedure adopted by engineers of assuming the kind 
bf reaction, and then calculating quantities and heats that should and would 
iccompany it, if it did take place and, however, crude this may look it is very 
useful, and in most cases good enough or rather as good as the knowledge of other 
Conditions to be met. 

Carbon and hydrogen are the only chemical elements of fuel value, and all 
Commercial fuels, including wood, peat, lignite, bituminous and anthracite 
ifcoals, charcoal, coke, crude petroleum oils with their distillates, gasolene, 
ikerosene, and their residues, tars, heavy oils, alcohols, benzole a bituminous 
tji'.oal product, natural gas as well as blast furnace, carburetted and uncar- 
'Jmretted water gas, coal gas, producer gas, and oil gas, are compounds, 
Mixtures and mixtures of compounds of these fuel elements, with oxygen 
'.la some cases. The one exception is sulphur, which exists in many solid, 
aquid and gaseous fuels as an impurity, but which also has some small 
. uel value. 

. i This being the case, the number of products to be formed by the complete 
jombustion of fuel is also small, and includes only carbon dioxide and water 
i ! apor, with the nitrogen carried by the air, and a small amount of sulphur 
iompounds, often ignored. 

The process of reaction, whether combustion or one of the related ones, 
I* to be described by the usual chemical equation which has the additional 
gnificance of showing the relative weights involved directly, because, 

(a) The total number of atoms of each chemical element on each side of 
the equation must be the same. 

(b) The sum of the products of the atomic weight of each atom and the 
number present, must be the same on each side of the equation. 

This is the same as saying, (a) that the total weight of hydrogen in the prod- 
3ts must be the same as the total weight of hydrogen in the original mixture 
id so also for the other elements, and (b) the total weight of the original mixture 
' reacting substances must be the same as that of the products. 

Natural fuels while sometimes consisting of simple elements like carbon 
hydrogen alone, or simple compounds like carbon monoxide or methane alone, 
'e more often mixtures. Their reaction equations are then to be derived 
om combinations of the equations representing the reactions of elements and 
compounds with oxygen or with each other. To facilitate this the following 
ble of some characteristic reactions of simple substances is inserted : 



510 



ENGINEERING THERMODYNAMICS 



SOME COMBUSTION AND RELATED REACTIONS OF FUEL ELEMENTS AND 

COMPOUNDS 



Combustion o 


f f uelel ements J c 

Product of 
partial car- 
bon combus- 
tion 


!H 2 -f-O 2 -2H 2 O ..... 




V A / 

(2) 
(3) 


2CO-|-O 2 2CO 2 . . . 












Paraffin series CnHjw+2: 
Methane CH 4 +2O 2 = CO 2 +2H 2 O . . . 


(4) 


Combustion 
of 
Chemical 
Compounds 


Hydrocarbons 


Ethane C S H 6 +-O 2 

J-* 1 X~ TT 1 / 


= 2CO 2 +3H 2 O . . (5) 
) O 2 = wCO 2 + (n + 1 )H 2 O (6) 

O 2 =2CO 2 +2H 2 O . . (7) 
2 =3CO 2 +3H 2 O . . (8) 


General C w H 2 n+ 2 + (^ 

Olefines or Ethylene series C re H 2w : 
Ethylene CzK 

Propylene C 3 H 6 +- 






General C K H 2n + -^<X = nCO 2 +nH 2 O . . 


(9) 




Alcohols 


Monatomic alcohols CnH2n+iOH: 
Methyl alcohol CH 3 OH+|o 2 = CO 2 +2H 2 O . . 
Ethyl alcohol C 2 H 5 OH+3O 2 =2CO 2 +3H 2 O. . 
General CwH^ +iOH + O 2 = nCO 2 + (n + 1 )H 2 O 


(10) 
(11) 

(12) 



Related Reactions, Incomplete Combustion, Decom- 
position, Dissociation. 



2C+0 2 

CO 2 +H 2 = 

CH4+CO+H 2 

C+CO 2 = 

CO+H 2 O = 

2CO 2 = 

2H 2 O = 

CH 4 

C 2 H 4 = 



= 2CO .... (13; 
= CO+H 2 O . . (14 
= C 2 H 4 +H 2 O. . (15 
= 2CO (16: 



:C0 2 +H 2 . 

= 2CO+O 2 . 
= 2H 2 +O 2 . 
= C+2H 2 . 
= C+CH 4 . 



2CH 4 
3C 2 H 2 = 



C 2 H 2 +3H 2 
C 2 H 4 +2C 2 H 2 



(17 
.(18 
.(19 
.(20 

.(21 
.(22, 
.(23 
. (24 



All these and any other reactions are characterized by definite weigh 
relations which are given directly by the reaction equation by introducing th 
weight of each element from a table of atomic weights, and for this purpos 
the nearest whole number is close enough. For example, the complete com 
bustion of hexane, CeHn, the main constituent of gasolene, to carbon dioxid 
and water, is given by, 



or 



C 6 H4+f02 = 6CO2+7H 2 



HEAT AND MATTER 5H 

This may be interpreted as follows, taking the atomic weight of C = 12 of 
H=l, and of O = 16. 



(6X12 = 72) Ibs. carbon 
+ (14 XI =.14) Ibs. hydrogen 
+ (9^X2X16 = 304) Ibs. oxygen 



make 



(6X12 = 72) Ibs. carbon 
+ (6X2X16 = 192) Ibs. oxygen 
+ (7X2X1 = 14) Ibs. hydrogen 
+ (7X16 = 112) Ibs. oxygen 



or 



, iLt^u =86) lbs ' i hexane \ make / (72+192 = 264 Ibs. carbon dioxide 
+ (304) Ibs. oxygen e \ +(14+112 = 126) Ibs. water vapor 

This general weight relation may be simplified somewhat by considering 
the weight of each one of the substances as unity, thus yielding the following 
different, but equivalent interpretations : 

^1 Ib. hexane+ Ibs. oxygen j make ( Ibs. car. diox.+i^ Ibs. water.) 

f 1 Ib. oxy. + Ibs. hexane J make ( Ibs. car. diox.+^ Ibs. water.) 

/ 12fi x result from com-, 

1 Ib. carb. diox.+ Ibs.water ) plete combus-(-^. i bs> hex.+^ Ibs. oxy. ) 
^o4 / tion of \*o4 264 * / 

2fi4 x result from com-, 

Ib. water+-^| Ibs. car. diox. ) plete combus-/^ [b s . hex.+^| Ibs. oxy. ) 
i^b / tion of . V 1 ^" l^o / 



|It might also be said from the same data that, 

1 Ib. hexane completely burned with oxygen yields 390 Ibs. products. 

1 Ib. oxygen will completely burn ^ Ibs. hexane and yield 390 Ibs. products. 

This particular example can be analyzed in still another way, yielding a general 
expression for the combustion of an analyzed fuel. 

Fuel analyses are reported in two ways: (a) Proximate, giving per cent of 
each independent compound, or separately determined constituent substance; 
(b) ultimate, giving the per cent of each chemical element. Applying this dis- 
tinction to the original mixture of hexane and oxygen, and its products the 
proximate analyses are: 



Original mixture 



Hexane = 390 

304 



Oxygen =^ = 77.9 % by weight. 

Total 100.00% 



512 



ENGINEERING THERMODYNAMICS 



Products 



264 
Carbon dioxide =5Hn 



67.6 % by weight. 



126 



Water vapor or liquid = ^ = 23.4 % by weight. 

Total 100.00% 

Using the ultimate analysis method of designation for mixtures and products 

72 



Original mixture 



Products 



Carbon (in hexane) 
Hydro, (in hexane) 

Oxygen (free) 
Total 

Carbon (in C0 2 ) 
Hydrogen (water) 
Oxygen (in C0 2 ) 

Oxygen (water) 

Total 



14 
390 

304 
: 390 



72 



14 
390 

192 
: 390 

112 
: 390 



= 18.4% by weight. 
= 3.6 % by weight. 

= 78.0 % by weight. 
100.00% 

= 18.4 % by weight. 
= 3.6 % by weight. 



-fjj- 78.0 % by weight. 



100.00% 



Neither combustion or its related reactions take place with oxygen alone, 
but with air containing oxygen, 23.2 per cent, and nitrogen, 76.8 per cent 

76 8 

by weight, each pound of oxygen carrying with it ^- = 3.31 Ibs. of nitrogen 

2i*j.2i 

or existing in 4.31 Ibs. of air. The nitrogen is generally considered neutral, 
though it may form compounds with hydrogen, such as ammonia directly, 
or with oxygen, such as nitrous oxide if conditions are right. If neutral, it 
has the effect of changing the weight of the mixture by an amount depending 
on the proportion of oxygen that came from air. 

/ (Weight of mixture with oxygen alone) \ / i . r . - 1 \ 

' +3.31 (wt. oxygen present) } = < wei e ht of mlxture wth alr) - 



or 



air is used instead of oxygen) 



- 3.31 (wt. of oxygen present). 



HEAT x\ND MATTER 



513 



This will add in the present case 304X3.31 = 1006 Ibs. of nitrogen in the 
combustion of one molecule or 86 Ibs. of hexane, the sum of 304 Ibs. of oxygen 
and 1006 Ibs. of nitrogen, giving 1310 Ibs., the weight of air required, and rais- 
ing the total weight of the mixture to 1396 Ibs. Of course, the per cent of the 
various constituents of the mixture and products is now changed, but the 
amount of change is quickly computed. All these relative numbers can be 
conveniently given in tables of conversion factors, such a table for hexane 
being given below, as a type form, useful in e very-day work when it is nec- 
essary to make repeated calculations with some one fuel. 

CONVERSION FACTORS FOR WEIGHTS IN THE COMPLETE COMBUSTION OF 
HEXANE WITH AIR TO WATER AND CARBON DIOXIDE 



Original 
! Mixture. 


Final 
Mixture. 


Hexane 
C 6 Hi4. 


Oxygen 


Nitrogen 
N. 


Air. 


Carbon Di- 
oxide CO 2 . 


Water 
HzO. 


1396. 


1396. 


86. 


304. 


1006. 


1310. 


264. 


126. 


1. 


1. 


.0616 


.218 


.720 .938 


.189 


.091 


16.22 


16.22 


1. 


3.54 


11.68 15.22 


3.07 


1.46 


4.59 


4.59 


.283 


1. 


3.31 4.31 


.87 


.41 


1.387 


1.387 


.085 


.302 


1. 


1.302 


.262 


.125 


1.066 


1.066 


.066 


.232 


.768 


1. 


.202 


.096 


5.29 


5.29 


.326 


1.15 


3.81 


4.96 


1. 


.48 


11.08 


11.08 


.683 


2.42 


7.98 


10.40 


2.10 


1. 



The average analysis of pure air is given by Parkes as being 02 = 20.96 per cent; 
|J0 2 = .04 per cent; N2 = 76.00 per cent by volume, giving 02 = 23.20 per cent; 
302 = .06 per cent; N2 = 76.74 per cent by weight. Regarding the C02 as 
Deing negligible, the relation may be used for the purpose of computations of 
ihis sort, O2 = 23.2 per cent and N2 = 76.8 per cent. On the assumption that 
my nitrogen present came from air and not from any other compound, such 
is ammonia, it must have been associated with oxygen in the proportion 3.31 
X) 1 of O2. The weight of oxygen on both sides must be equal and the weight 
)f air on one side must be 4.31 times the weight of nitrogen. Of the nitrogen 

,hat is present X(the weight of nitrogen present) must have come from air 

o.ol 

tnd the rest from the other substances. 

A much-used weight relation is, for the weight of air and products per pound 
)f fuel, and on the assumption that 

H = part of hydrogen by weight per pound fuel, 
C = part of carbon by weight per pound fuel, 

00 

Weight of air per Ib. of fuel)=CXY|x4,31+HX8X4.31 = 11.49C+34.46H,(7i:) 



514 ENGINEERING THERMODYNAMICS 

32 
Weight of products per Ib. fuel = 3|C Ibs. C0 2 +9H Ibs. H 2 O+^ 

X3.31C Ibs. N+8X3.31H Ibs. N 

= 3.67C in form of CO 2 +9H in form of H 2 O+8.82 C in form of N 
= 12.49C+35.46H, ................ (715 

for complete combustion in air with no more air supplied than enters into th< 
reaction 

Most of the practical problems concerning the relative amounts of substance; 
involved in combustion and reduction processes are concerned with gases, a 
least on one side of the equation and sometimes on both. Direct combustioi 
and the gasification of fuels in producers and complete combustion in boilei 
furnaces always yields gas mixtures, the composition of which is always volui 
metrically determined by analysis, while the explosive mixtures or priman 
working substances of gas engines are gaseous initially and remain so afte:j 
combustion. It is quite necessary, therefore, to transform the weight relation; 
of the reaction equation into another form yielding volumes. There are thre 
ways of doing this, all equivalent and all yielding the same result if the constant' 
used are consistent, and if the gases and vapors follow the Avagadro law, oi 
which the most useful of the methods depends. 

1. The volume at standard conditions of any substance can be found fron 
the weight present by multiplying that weight by the specific volume of th' 
substances ; in cubic feet per pound, at standard condition of 1 atm. at 32 F. 

2. The molecular weight in pounds of any gaseous or vapor substanc 
assumed to follow Avagadro's law occupies 358 cu.ft. 

3. The volumetric relations are given directly by the coefficients of the sub 
stance in the chemical equation when the .substances are gaseous and all ente 
into the reaction, and when each is expressed in terms of molecules present. 

The first method needs no explanation or further development and the secom 
and third are really one. // in the reaction equations there be written m 
any substance its molecular weightX358, the product mil be the volume c< 
that substance in the reaction in cubic fe